[proofplan] We prove the cycle of implications $(1) \Rightarrow (3) \Rightarrow (2) \Rightarrow (1)$. Compactness implies sequential compactness via a countable-base argument: given a sequence, the second-countable topology provides a nested sequence of basic open sets that pins down a convergent subsequence. Sequential compactness implies countable compactness because any countable cover without a finite subcover yields a sequence with no convergent subsequence (the argument of [Sequential Compactness Plus Lindelof Implies Compactness](/theorems/1043)). Finally, countable compactness implies compactness because every second-countable space is Lindel\"{o}f, reducing arbitrary covers to countable ones. [/proofplan]

[step:$(1) \Rightarrow (3)$: Extract a convergent subsequence from any sequence in a compact second-countable space]Let $X$ be a compact second-countable topological space, and let $\{x_n\}_{n=1}^\infty$ be an arbitrary sequence in $X$. Let $\mathcal{B} = \{B_k\}_{k=1}^\infty$ be a countable base for the topology of $X$. We construct a convergent subsequence by a diagonal argument on the base elements. Define $S_0 := \mathbb{N}$. For each $k \in \mathbb{N}$, consider the base element $B_k$. The set $\{n \in S_{k-1} : x_n \in B_k\}$ is either infinite or has a finite complement in $S_{k-1}$ (relative to elements indexing points in $B_k$). Precisely, at least one of the two sets \begin{align*} S_{k-1}^+ &:= \{n \in S_{k-1} : x_n \in B_k\}, \\ S_{k-1}^- &:= \{n \in S_{k-1} : x_n \notin B_k\} \end{align*} is infinite. Define $S_k$ to be an infinite one (choosing $S_{k-1}^+$ if both are infinite). This produces a nested chain $S_0 \supset S_1 \supset S_2 \supset \cdots$ of infinite subsets of $\mathbb{N}$ with the property that for each $k$, either $x_n \in B_k$ for all $n \in S_k$, or $x_n \notin B_k$ for all $n \in S_k$. Extract the diagonal subsequence inductively: let $n_1$ be the smallest element of $S_1$, and for $k \ge 2$, let $n_k$ be the smallest element of $S_k$ that exceeds $n_{k-1}$ (such an element exists because $S_k$ is infinite). The sequence $\{n_k\}_{k=1}^\infty$ is strictly increasing by construction, and for every $j \le k$, we have $n_k \in S_k \subset S_j$, so the subsequence $\{x_{n_k}\}_{k \ge j}$ is eventually contained in $S_j$.[/step]

[guided]Let $X$ be a compact second-countable topological space, and let $\{x_n\}_{n=1}^\infty$ be a sequence in $X$. Let $\mathcal{B} = \{B_k\}_{k=1}^\infty$ be a countable base for the topology of $X$. The strategy is to thin the index set of the sequence repeatedly — once for each base element — so that the surviving subsequence has a definite relationship (always in, or always out) with every basic open set. This rigidity will force convergence. Define $S_0 := \mathbb{N}$, the full index set. At stage $k \ge 1$, consider the base element $B_k$ and the current infinite index set $S_{k-1}$. Partition $S_{k-1}$ according to membership in $B_k$: \begin{align*} S_{k-1}^+ &:= \{n \in S_{k-1} : x_n \in B_k\}, \\ S_{k-1}^- &:= \{n \in S_{k-1} : x_n \notin B_k\}. \end{align*} Since $S_{k-1} = S_{k-1}^+ \cup S_{k-1}^-$ and $S_{k-1}$ is infinite, at least one of $S_{k-1}^+$, $S_{k-1}^-$ is infinite. Define $S_k$ to be an infinite one (choosing $S_{k-1}^+$ if both are infinite). The key property is: for each $k$, the set $S_k$ is infinite, and either $\{x_n : n \in S_k\} \subset B_k$ or $\{x_n : n \in S_k\} \subset X \setminus B_k$. This gives a nested chain $S_0 \supset S_1 \supset S_2 \supset \cdots$ of infinite subsets of $\mathbb{N}$. We extract a diagonal subsequence inductively: let $n_1$ be the smallest element of $S_1$, and for $k \ge 2$, let $n_k$ be the smallest element of $S_k$ that exceeds $n_{k-1}$ (such an element exists because $S_k$ is infinite). By construction, $n_1 < n_2 < n_3 < \cdots$. For every fixed $j$ and every $k \ge j$, the nesting $S_k \subset S_j$ gives $n_k \in S_j$, so the diagonal subsequence is eventually contained in $S_j$ and hence eventually agrees with $S_j$ in its relationship to $B_j$.[/guided]

[step:Show the diagonal subsequence converges by identifying its limit via compactness]We claim the subsequence $\{x_{n_k}\}_{k=1}^\infty$ has a cluster point, and that this cluster point is in fact a limit. Since $X$ is compact, the collection of closed sets $\{F_j\}_{j=1}^\infty$ defined by $F_j := \overline{\{x_{n_k} : k \ge j\}}$ satisfies the finite intersection property: any finite subcollection $F_{j_1}, \ldots, F_{j_r}$ satisfies $F_{\max(j_1,\ldots,j_r)} \subset F_{j_1} \cap \cdots \cap F_{j_r}$, and each $F_j$ is nonempty (as $\{x_{n_k}\}_{k \ge j}$ is an infinite set). By compactness, $\bigcap_{j=1}^\infty F_j \neq \varnothing$. Pick $x \in \bigcap_{j=1}^\infty F_j$. Now let $U$ be any open set containing $x$. Since $\mathcal{B}$ is a base, there exists $B_m \in \mathcal{B}$ with $x \in B_m \subset U$. Because $x \in F_m = \overline{\{x_{n_k} : k \ge m\}}$, every open set containing $x$ meets $\{x_{n_k} : k \ge m\}$, so in particular $B_m$ contains $x_{n_k}$ for some $k \ge m$. But by the diagonal construction, for all $k \ge m$ we have $n_k \in S_m$, and $S_m$ was chosen so that either all its indexed points lie in $B_m$ or none do. Since at least one $x_{n_k}$ with $k \ge m$ lies in $B_m$, the "all in" alternative holds: $x_{n_k} \in B_m \subset U$ for all $k \ge m$. Since $U$ was an arbitrary open set containing $x$, the subsequence $x_{n_k} \to x$. This proves $X$ is sequentially compact.[/step]

[guided]We now show the diagonal subsequence converges. Define the closed sets $F_j := \overline{\{x_{n_k} : k \ge j\}}$ for each $j \in \mathbb{N}$. These sets are nested: $F_1 \supset F_2 \supset \cdots$, and each is nonempty and closed. Any finite subcollection has nonempty intersection (it contains the set with the largest index), so $\{F_j\}_{j=1}^\infty$ has the finite intersection property. Since $X$ is compact, every family of closed sets with the finite intersection property has nonempty total intersection. We select a point $x \in \bigcap_{j=1}^\infty F_j$. We claim $x_{n_k} \to x$. Let $U$ be any open neighbourhood of $x$. Since $\mathcal{B}$ is a base, there exists $B_m \in \mathcal{B}$ with $x \in B_m \subset U$. We need to show $x_{n_k} \in U$ for all sufficiently large $k$. Since $x \in F_m = \overline{\{x_{n_k} : k \ge m\}}$, the point $x$ lies in the closure of the $m$-tail of the subsequence. In particular, the open set $B_m$ (which contains $x$) must intersect $\{x_{n_k} : k \ge m\}$, so there exists at least one index $k_0 \ge m$ with $x_{n_{k_0}} \in B_m$. Now we invoke the diagonal construction. For all $k \ge m$, the index $n_k$ belongs to $S_m$ (by the nesting $S_k \subset S_m$ for $k \ge m$). By the construction of $S_m$, either $x_n \in B_m$ for all $n \in S_m$, or $x_n \notin B_m$ for all $n \in S_m$. Since we just found $x_{n_{k_0}} \in B_m$ with $n_{k_0} \in S_m$, the second alternative is excluded. Therefore $x_{n_k} \in B_m \subset U$ for all $k \ge m$. Since $U$ was an arbitrary open neighbourhood of $x$, we conclude $x_{n_k} \to x$.[/guided]

[step:$(3) \Rightarrow (2)$: Sequential compactness implies countable compactness]Let $\{V_n\}_{n=1}^\infty$ be a countable open cover of $X$. We must show it has a finite subcover. Suppose for contradiction that no finite subcollection covers $X$. Then for every $N \in \mathbb{N}$, there exists $x_N \in X \setminus \bigcup_{n=1}^N V_n$. By sequential compactness, the sequence $\{x_N\}_{N=1}^\infty$ has a convergent subsequence $x_{N_k} \to x$ for some $x \in X$. Since $\{V_n\}_{n=1}^\infty$ covers $X$, there exists $m \in \mathbb{N}$ with $x \in V_m$. The set $V_m$ is open, so $x_{N_k} \in V_m$ for all sufficiently large $k$. But for $k$ large enough that $N_k \ge m$, the construction gives $x_{N_k} \notin V_n$ for all $n \le N_k$, and in particular $x_{N_k} \notin V_m$ — a contradiction. This is precisely the argument of [Sequential Compactness Plus Lindelof Implies Compactness](/theorems/1043) applied to a countable cover directly, without needing the Lindel\"{o}f hypothesis (since the cover is already countable).[/step]

[guided]We show that every countable open cover of a sequentially compact space has a finite subcover. Let $\{V_n\}_{n=1}^\infty$ be a countable open cover of $X$. Assume for contradiction that no finite subcollection covers $X$. Then for each $N \in \mathbb{N}$, the set $X \setminus \bigcup_{n=1}^N V_n$ is nonempty, and we choose \begin{align*} x_N \in X \setminus \bigcup_{n=1}^N V_n. \end{align*} The defining property is $x_N \notin V_n$ for all $n \le N$. By sequential compactness, there exists a subsequence $\{x_{N_k}\}_{k=1}^\infty$ converging to some $x \in X$. Since $\{V_n\}_{n=1}^\infty$ covers $X$, there exists $m \in \mathbb{N}$ with $x \in V_m$. Because $V_m$ is open and $x_{N_k} \to x$, there exists $k_0$ with $x_{N_k} \in V_m$ for all $k \ge k_0$. Since $N_k \to \infty$, there exists $k_1$ with $N_{k_1} \ge m$. Setting $k^* = \max\{k_0, k_1\}$, we have both $x_{N_{k^*}} \in V_m$ and $x_{N_{k^*}} \notin V_m$ (the latter because $N_{k^*} \ge m$ and $x_{N_{k^*}} \notin V_n$ for all $n \le N_{k^*}$). This is a contradiction, so $\{V_n\}_{n=1}^\infty$ admits a finite subcover. Note that this step does not use the second-countable hypothesis — it is a general fact about sequentially compact spaces and countable covers. The connection to theorem 1043 is worth recording: theorem 1043 applies this same argument after first using the Lindel\"{o}f property to reduce an arbitrary cover to a countable one. Here, the cover is already countable, so no reduction is needed.[/guided]

[step:$(2) \Rightarrow (1)$: Countable compactness plus the Lindel\"{o}f property yields compactness]Let $\{U_\alpha\}_{\alpha \in A}$ be an arbitrary open cover of $X$. Since $X$ is second-countable, $X$ is Lindel\"{o}f: every open cover has a countable subcover. To verify this, let $\mathcal{B} = \{B_k\}_{k=1}^\infty$ be a countable base. For each $\alpha \in A$ and each $x \in U_\alpha$, there exists $B_{k(x,\alpha)} \in \mathcal{B}$ with $x \in B_{k(x,\alpha)} \subset U_\alpha$. Define $\mathcal{B}' := \{B_k \in \mathcal{B} : B_k \subset U_\alpha \text{ for some } \alpha \in A\}$. Then $\mathcal{B}'$ is a countable collection of basic open sets whose union equals $X$. For each $B_k \in \mathcal{B}'$, choose one $\alpha(k) \in A$ with $B_k \subset U_{\alpha(k)}$. Then $\{U_{\alpha(k)}\}_{B_k \in \mathcal{B}'}$ is a countable subcover of the original cover. By countable compactness, this countable subcover has a finite subcover. Therefore $X$ is compact.[/step]

[guided]Let $\{U_\alpha\}_{\alpha \in A}$ be an arbitrary open cover of $X$. We need a finite subcover. The key observation is that second countability implies the Lindel\"{o}f property. We verify this directly. Let $\mathcal{B} = \{B_k\}_{k=1}^\infty$ be a countable base for $X$. For each point $x \in X$, there exists some $\alpha \in A$ with $x \in U_\alpha$. Since $\mathcal{B}$ is a base, there exists $B_k \in \mathcal{B}$ with $x \in B_k \subset U_\alpha$. Define the subcollection \begin{align*} \mathcal{B}' := \{B_k \in \mathcal{B} : B_k \subset U_\alpha \text{ for some } \alpha \in A\}. \end{align*} This subcollection covers $X$: for every $x \in X$, we just showed there exists $B_k \in \mathcal{B}'$ containing $x$. Moreover, $\mathcal{B}'$ is countable (as a subset of the countable collection $\mathcal{B}$). For each $B_k \in \mathcal{B}'$, select one index $\alpha(k) \in A$ with $B_k \subset U_{\alpha(k)}$ (using the axiom of choice for a countable collection). Then \begin{align*} X = \bigcup_{B_k \in \mathcal{B}'} B_k \subset \bigcup_{B_k \in \mathcal{B}'} U_{\alpha(k)}, \end{align*} so $\{U_{\alpha(k)}\}_{B_k \in \mathcal{B}'}$ is a countable subcover of the original cover. By countable compactness (hypothesis (2)), every countable open cover admits a finite subcover. Applying this to $\{U_{\alpha(k)}\}_{B_k \in \mathcal{B}'}$ yields the desired finite subcover. Since the original open cover was arbitrary, $X$ is compact. This step is the only place in the cycle where the second-countable hypothesis is consumed for the $(2) \Rightarrow (1)$ direction. Without second countability, countable compactness does not imply compactness: the first uncountable ordinal $\omega_1$ with the order topology is countably compact but not compact (and not second-countable).[/guided]