[proofplan] The forward direction is immediate (a [compact space](/page/Compact%20Space) satisfies the finite subcover condition for every open cover, hence in particular for subbasis covers). For the converse, we assume that every subbasis cover has a finite subcover and prove compactness. Suppose for contradiction that there exists an open cover $\mathcal{C}$ with no finite subcover. We apply Zorn's Lemma to extend $\mathcal{C}$ to a maximal open cover $\mathcal{M}$ with no finite subcover. The maximality of $\mathcal{M}$ forces every point of $X$ to lie in some subbasis element outside $\mathcal{M}$, and a careful analysis of the subbasis intersection structure yields a finite subcover from the subbasis — contradicting the assumption on $\mathcal{M}$. [/proofplan]

[step:The forward direction is immediate from the definition of compactness] Assume $X$ is [compact](/page/Compact%20Space) and let $\mathcal{S}$ be a subbasis for the [topology](/page/Topology) on $X$. Every cover of $X$ by elements of $\mathcal{S}$ is in particular an open cover, so it admits a finite subcover by compactness. [/step]

[step:Assume the subbasis condition and suppose for contradiction that $X$ is not compact]Let $\mathcal{S}$ be a subbasis for the [topology](/page/Topology) on $X$ such that every cover of $X$ by members of $\mathcal{S}$ has a finite subcover. Suppose for contradiction that $X$ is not [compact](/page/Compact%20Space): there exists an open cover $\mathcal{C}_0$ of $X$ with no finite subcover.[/step]

[guided]We are proving the harder direction: subbasis finite subcover property implies compactness. The contrapositive assumption gives us a "bad" open cover $\mathcal{C}_0$ — one that covers $X$ but from which no finite subfamily suffices. The plan is to enlarge $\mathcal{C}_0$ to a *maximal* bad cover using Zorn's Lemma, then exploit maximality to derive a contradiction with the subbasis hypothesis.[/guided]

[step:Apply Zorn's Lemma to obtain a maximal open cover with no finite subcover]Let $\mathscr{P}$ denote the collection of all open covers of $X$ that contain $\mathcal{C}_0$ and have no finite subcover: \begin{align*} \mathscr{P} = \bigl\{\mathcal{C} : \mathcal{C} \text{ is an open cover of } X,\; \mathcal{C}_0 \subset \mathcal{C},\; \text{no finite subfamily of } \mathcal{C} \text{ covers } X\bigr\}. \end{align*} We partially order $\mathscr{P}$ by inclusion. The set $\mathscr{P}$ is non-empty since $\mathcal{C}_0 \in \mathscr{P}$. [claim:ChainsInPHaveUpperBounds] Every chain in $(\mathscr{P}, \subset)$ has an upper bound in $\mathscr{P}$. [/claim] [proof] Let $\{\mathcal{C}_\gamma\}_{\gamma \in \Gamma}$ be a chain in $\mathscr{P}$, and set $\mathcal{C}^* = \bigcup_{\gamma \in \Gamma} \mathcal{C}_\gamma$. Each element of $\mathcal{C}^*$ is an open subset of $X$ (since each $\mathcal{C}_\gamma$ consists of [open sets](/page/Open%20Set)), and $\mathcal{C}_0 \subset \mathcal{C}_\gamma \subset \mathcal{C}^*$ for all $\gamma$, so $\mathcal{C}^*$ is an open cover of $X$ containing $\mathcal{C}_0$. Suppose $\mathcal{C}^*$ had a finite subcover $\{U_1, \ldots, U_m\}$. Each $U_i$ belongs to some $\mathcal{C}_{\gamma_i}$. Since $\{\mathcal{C}_\gamma\}$ is a chain, the finite collection $\mathcal{C}_{\gamma_1}, \ldots, \mathcal{C}_{\gamma_m}$ has a maximum element $\mathcal{C}_{\gamma_j}$ (the one containing all others). Then $\{U_1, \ldots, U_m\} \subset \mathcal{C}_{\gamma_j}$, so $\mathcal{C}_{\gamma_j}$ has a finite subcover — contradicting $\mathcal{C}_{\gamma_j} \in \mathscr{P}$. Therefore $\mathcal{C}^*$ has no finite subcover, so $\mathcal{C}^* \in \mathscr{P}$. [/proof] By Zorn's Lemma, $\mathscr{P}$ contains a maximal element $\mathcal{M}$. The cover $\mathcal{M}$ is an open cover of $X$ that contains $\mathcal{C}_0$, has no finite subcover, and is maximal with respect to these properties.[/step]

[guided]Zorn's Lemma requires that every chain has an upper bound. The verification is standard: the union of a chain of covers is a cover (since the original $\mathcal{C}_0$ is contained in each element of the chain), and the union cannot have a finite subcover because any finite subcover would lie within a single element of the chain (by the chain condition), contradicting membership in $\mathscr{P}$. The maximality of $\mathcal{M}$ is the engine of the proof. It means: if we add any open set $U \notin \mathcal{M}$ to $\mathcal{M}$, the resulting cover $\mathcal{M} \cup \{U\}$ *does* have a finite subcover. We will exploit this rigidity in the next step.[/guided]

[step:Extract the key consequence of maximality: every open set outside $\mathcal{M}$ is "almost covered" by finitely many sets in $\mathcal{M}$][claim:MaximalityConsequence] For every [open set](/page/Open%20Set) $U \subset X$ with $U \notin \mathcal{M}$, there exist finitely many sets $M_1, \ldots, M_r \in \mathcal{M}$ such that $X = U \cup M_1 \cup \cdots \cup M_r$. [/claim] [proof] Since $\mathcal{M}$ is maximal in $\mathscr{P}$ and $U \notin \mathcal{M}$, the collection $\mathcal{M} \cup \{U\}$ is strictly larger than $\mathcal{M}$. By maximality, $\mathcal{M} \cup \{U\} \notin \mathscr{P}$. Since $\mathcal{M} \cup \{U\}$ is still an open cover of $X$ containing $\mathcal{C}_0$, the only way it can fail to be in $\mathscr{P}$ is by possessing a finite subcover. Let $\{V_1, \ldots, V_s\}$ be such a finite subcover. If $U \notin \{V_1, \ldots, V_s\}$, then $\{V_1, \ldots, V_s\} \subset \mathcal{M}$ is a finite subcover from $\mathcal{M}$, contradicting $\mathcal{M} \in \mathscr{P}$. Therefore $U \in \{V_1, \ldots, V_s\}$. Setting $\{M_1, \ldots, M_r\} = \{V_1, \ldots, V_s\} \setminus \{U\}$, we obtain $X = U \cup M_1 \cup \cdots \cup M_r$ with each $M_i \in \mathcal{M}$. [/proof][/step]

[guided]This claim is the critical structural fact. In words: $\mathcal{M}$ is so large that adding any single open set $U$ to it would create a finite subcover. So $U$ together with finitely many members of $\mathcal{M}$ already cover $X$, which means $X \setminus U$ is covered by finitely many members of $\mathcal{M}$. We will use this in the next step to produce a subbasis cover of $X$ with no finite subcover, contradicting the hypothesis.[/guided]

[step:Show that $\mathcal{M}$ cannot contain every subbasis element that covers a given point]We claim that there exists a point $x \in X$ such that no subbasis element $S \in \mathcal{S}$ containing $x$ belongs to $\mathcal{M}$. Suppose the contrary: for every $x \in X$, there exists $S_x \in \mathcal{S} \cap \mathcal{M}$ with $x \in S_x$. Then $\{S_x\}_{x \in X}$ is a cover of $X$ by members of the subbasis $\mathcal{S}$. By the subbasis hypothesis, this cover has a finite subcover $\{S_{x_1}, \ldots, S_{x_m}\}$. But each $S_{x_i} \in \mathcal{M}$, so $\{S_{x_1}, \ldots, S_{x_m}\}$ is a finite subcover of $X$ from $\mathcal{M}$ — contradicting $\mathcal{M} \in \mathscr{P}$. Therefore there exists $x_0 \in X$ such that for every $S \in \mathcal{S}$ with $x_0 \in S$, we have $S \notin \mathcal{M}$.[/step]

[guided]This step is where the subbasis hypothesis is consumed. The argument is by contradiction within the contradiction: we already assumed $X$ is not [compact](/page/Compact%20Space), which gave us $\mathcal{M}$. Now we test whether $\mathcal{M}$ is compatible with the subbasis hypothesis. If every point were covered by some subbasis element in $\mathcal{M}$, the subbasis hypothesis would yield a finite subcover from $\mathcal{M}$, which is impossible. So there must exist a "stubborn" point $x_0$ that is not covered by any subbasis element belonging to $\mathcal{M}$. This means every subbasis element containing $x_0$ lies *outside* $\mathcal{M}$, and we can apply the maximality consequence to each of them.[/guided]

[step:Derive the final contradiction using the subbasis intersection structure]Since $\mathcal{M}$ is an open cover of $X$, there exists $W \in \mathcal{M}$ with $x_0 \in W$. Since $\mathcal{S}$ is a subbasis for the [topology](/page/Topology) on $X$ and $W$ is open, there exist $S_1, \ldots, S_n \in \mathcal{S}$ such that \begin{align*} x_0 \in S_1 \cap S_2 \cap \cdots \cap S_n \subset W. \end{align*} (Every [open set](/page/Open%20Set) containing $x_0$ contains a basic open set — a finite intersection of subbasis elements — around $x_0$.) By the conclusion of the previous step, each $S_i \notin \mathcal{M}$ (since $x_0 \in S_i$ and $S_i \in \mathcal{S}$). By the maximality consequence (the claim in the earlier step), for each $i = 1, \ldots, n$, there exist finitely many sets $M_{i,1}, \ldots, M_{i,r_i} \in \mathcal{M}$ such that \begin{align*} X = S_i \cup M_{i,1} \cup \cdots \cup M_{i,r_i}. \end{align*} Equivalently, $X \setminus S_i \subset M_{i,1} \cup \cdots \cup M_{i,r_i}$, so any point not in $S_i$ is covered by finitely many members of $\mathcal{M}$. Now consider any point $y \in X$. Either $y \in S_1 \cap \cdots \cap S_n$, in which case $y \in W$ and $W \in \mathcal{M}$; or $y \notin S_i$ for some $i \in \{1, \ldots, n\}$, in which case $y \in M_{i,1} \cup \cdots \cup M_{i,r_i}$. Therefore \begin{align*} X = W \cup \bigcup_{i=1}^n \bigcup_{j=1}^{r_i} M_{i,j}. \end{align*} This is a finite subcover of $X$ from $\mathcal{M}$ (the set $W$ and the finitely many sets $M_{i,j}$ all belong to $\mathcal{M}$), contradicting the fact that $\mathcal{M} \in \mathscr{P}$ has no finite subcover. This contradiction shows that the assumption "$X$ is not [compact](/page/Compact%20Space)" is false.[/step]

[guided]This is the heart of the argument. We have a point $x_0$ with the property that every subbasis element containing $x_0$ lies outside $\mathcal{M}$. But $x_0$ must be covered by *some* member $W$ of $\mathcal{M}$. Since $W$ is open and $\mathcal{S}$ is a subbasis, the set $W$ contains a finite subbasis intersection around $x_0$: $x_0 \in S_1 \cap \cdots \cap S_n \subset W$. Each $S_i$ is outside $\mathcal{M}$, so by the maximality consequence, the complement $X \setminus S_i$ is covered by finitely many members of $\mathcal{M}$. The key idea is that the complement of the intersection $S_1 \cap \cdots \cap S_n$ is the union $\bigcup_{i=1}^n (X \setminus S_i)$ (by De Morgan's law), and each piece $X \setminus S_i$ requires only finitely many members of $\mathcal{M}$. So the complement of $S_1 \cap \cdots \cap S_n$ is covered by finitely many members of $\mathcal{M}$, and the intersection $S_1 \cap \cdots \cap S_n$ itself is contained in $W \in \mathcal{M}$. Together, finitely many members of $\mathcal{M}$ cover all of $X$ — a contradiction. The finiteness of the intersection $S_1 \cap \cdots \cap S_n$ is essential: if we needed an infinite intersection of subbasis elements to fit inside $W$, the De Morgan complement would be an infinite union of sets, each requiring finitely many members of $\mathcal{M}$, and the total would no longer be finite.[/guided]