[proofplan] The argument proceeds in two stages. First, we reduce an arbitrary open cover to a countable one using the Lindel\"{o}f hypothesis. Second, we show that a countable open cover of a sequentially compact space admits a finite subcover by contraposition: if no finite subcollection covers $X$, we construct a sequence with no convergent subsequence, contradicting sequential compactness. [/proofplan]

[step:Reduce to a countable open cover via the Lindel\"{o}f property]Let $\{U_\alpha\}_{\alpha \in A}$ be an arbitrary open cover of $X$. Because $X$ is Lindel\"{o}f, there exists a countable subset $\{\alpha_n\}_{n=1}^\infty \subset A$ such that $X = \bigcup_{n=1}^\infty U_{\alpha_n}$. It therefore suffices to show that every countable open cover of $X$ admits a finite subcover.[/step]

[guided]Let $\{U_\alpha\}_{\alpha \in A}$ be an arbitrary open cover of $X$, where the index set $A$ is potentially uncountable. The Lindel\"{o}f property asserts that every open cover of $X$ has a countable subcover. Applying this directly, we extract a countable subcollection $\{U_{\alpha_n}\}_{n=1}^\infty$ that still covers $X$: \begin{align*} X = \bigcup_{n=1}^\infty U_{\alpha_n}. \end{align*} The problem of finding a finite subcover from the original family $\{U_\alpha\}_{\alpha \in A}$ thus reduces to the problem of finding a finite subcover from the countable family $\{U_{\alpha_n}\}_{n=1}^\infty$. This is the essential role of the Lindel\"{o}f hypothesis: it bridges the gap between arbitrary open covers (which sequential compactness says nothing about directly) and countable open covers (which sequential compactness can control).[/guided]

[step:Show that a countable open cover of a sequentially compact space has a finite subcover]Relabel the countable cover as $\{V_n\}_{n=1}^\infty$. Suppose for contradiction that no finite subcollection covers $X$. Then for every $N \in \mathbb{N}$, the partial union $\bigcup_{n=1}^N V_n$ does not cover $X$, so there exists a point \begin{align*} x_N \in X \setminus \bigcup_{n=1}^N V_n. \end{align*} This defines a sequence $\{x_N\}_{N=1}^\infty$ in $X$. Because $X$ is sequentially compact, this sequence has a subsequence $\{x_{N_k}\}_{k=1}^\infty$ converging to some limit $x \in X$.[/step]

[guided]Relabel the countable subcover as $\{V_n\}_{n=1}^\infty$ with $X = \bigcup_{n=1}^\infty V_n$. Suppose for the sake of contradiction that this cover admits no finite subcover. Then for every $N \in \mathbb{N}$, the partial union $\bigcup_{n=1}^N V_n$ fails to cover $X$, so the complement $X \setminus \bigcup_{n=1}^N V_n$ is nonempty. We select a point \begin{align*} x_N \in X \setminus \bigcup_{n=1}^N V_n. \end{align*} The defining property of $x_N$ is that $x_N \notin V_n$ for all $n \le N$. This construction produces a sequence $\{x_N\}_{N=1}^\infty$ in $X$. By the sequential compactness of $X$, there exists a subsequence $\{x_{N_k}\}_{k=1}^\infty$ and a point $x \in X$ such that $x_{N_k} \to x$ as $k \to \infty$. We now show this convergence leads to a contradiction.[/guided]

[step:Derive a contradiction from the convergence of the subsequence]Since $\{V_n\}_{n=1}^\infty$ covers $X$, there exists an index $m \in \mathbb{N}$ such that $x \in V_m$. The set $V_m$ is open, so because $x_{N_k} \to x$, there exists $k_0 \in \mathbb{N}$ such that $x_{N_k} \in V_m$ for all $k \ge k_0$. Since $\{N_k\}_{k=1}^\infty$ is a strictly increasing sequence of natural numbers, we may choose $k$ large enough so that $N_k \ge m$. For such $k$, the construction in the previous step guarantees $x_{N_k} \notin V_n$ for all $n \le N_k$. In particular, since $m \le N_k$, we have $x_{N_k} \notin V_m$. This contradicts $x_{N_k} \in V_m$. The contradiction shows that the assumption — that no finite subcollection of $\{V_n\}_{n=1}^\infty$ covers $X$ — is false. Therefore $\{V_n\}_{n=1}^\infty$, and hence the original cover $\{U_\alpha\}_{\alpha \in A}$, admits a finite subcover.[/step]

[guided]Since $\{V_n\}_{n=1}^\infty$ covers $X$ and $x \in X$, there exists an index $m \in \mathbb{N}$ with $x \in V_m$. The set $V_m$ is open in $X$, and $x_{N_k} \to x$, so there exists $k_0 \in \mathbb{N}$ such that \begin{align*} x_{N_k} \in V_m \quad \text{for all } k \ge k_0. \end{align*} On the other hand, the subsequence indices $\{N_k\}_{k=1}^\infty$ form a strictly increasing sequence of natural numbers, so $N_k \to \infty$ as $k \to \infty$. In particular, there exists $k_1 \in \mathbb{N}$ with $N_{k_1} \ge m$. Set $k^* = \max\{k_0, k_1\}$. Then $N_{k^*} \ge m$, and by the construction of the sequence, $x_{N_{k^*}} \notin V_n$ for all $n \le N_{k^*}$. Since $m \le N_{k^*}$, this gives $x_{N_{k^*}} \notin V_m$. But $k^* \ge k_0$ also gives $x_{N_{k^*}} \in V_m$. This is a contradiction. Since the assumption that no finite subcollection covers $X$ leads to a contradiction, we conclude that $\{V_n\}_{n=1}^\infty$ admits a finite subcover. Since $\{V_n\}_{n=1}^\infty$ is itself a subcollection of the original cover $\{U_\alpha\}_{\alpha \in A}$, this finite subcover also serves as a finite subcover of the original cover. Therefore $X$ is compact.[/guided]