[proofplan] Given an arbitrary open cover $\{U_\alpha\}_{\alpha \in A}$ of $X$, we use the countable base $\mathcal{B}$ to reduce the problem to countable combinatorics. Each point of $X$ lies in some $U_\alpha$, and the base property provides a basis element sandwiched between the point and $U_\alpha$. The set of basis element indices that arise is a countable subset of $\mathbb{N}$, and selecting one $U_\alpha$ per such index yields a countable subcover. [/proofplan]

[step:Assign a basis element to each point to reduce the cover to a countable index set]Let $\mathcal{B} = \{B_n\}_{n=1}^\infty$ be a countable base for $\tau$, and let $\{U_\alpha\}_{\alpha \in A}$ be an open cover of $X$. For each $x \in X$, choose $\alpha(x) \in A$ with $x \in U_{\alpha(x)}$. Since $\mathcal{B}$ is a base and $U_{\alpha(x)}$ is open, there exists $n(x) \in \mathbb{N}$ with \begin{align*} x \in B_{n(x)} \subset U_{\alpha(x)}. \end{align*} Define $S := \{n(x) : x \in X\} \subset \mathbb{N}$. Since $S$ is a subset of $\mathbb{N}$, it is countable.[/step]

[guided]Let $\mathcal{B} = \{B_n\}_{n=1}^\infty$ be a countable base for $\tau$, and let $\{U_\alpha\}_{\alpha \in A}$ be an open cover of $X$, so that $X = \bigcup_{\alpha \in A} U_\alpha$. The index set $A$ may be uncountable --- the task is to extract a countable subfamily that still covers $X$. The strategy is to pass from the potentially uncountable indexing by $A$ to the countable indexing by $\mathbb{N}$, using the base as a "bottleneck." For each $x \in X$, choose some $\alpha(x) \in A$ with $x \in U_{\alpha(x)}$. Since $U_{\alpha(x)}$ is open and $\mathcal{B}$ is a base, there exists an index $n(x) \in \mathbb{N}$ such that \begin{align*} x \in B_{n(x)} \subset U_{\alpha(x)}. \end{align*} The set $S := \{n(x) : x \in X\} \subset \mathbb{N}$ records which basis elements appeared. Since $S \subset \mathbb{N}$, it is countable (finite or countably infinite). This is the key reduction: we have replaced the uncountable family $\{U_\alpha\}$ with the countable family $\{B_n\}_{n \in S}$ of basis elements.[/guided]

[step:Select one covering set per basis index to produce a countable subcover]For each $n \in S$, choose one index $\alpha_n \in A$ such that $B_n \subset U_{\alpha_n}$. (Such an $\alpha_n$ exists because $n \in S$ means $n = n(x)$ for some $x$, and we had $B_{n(x)} \subset U_{\alpha(x)}$; set $\alpha_n := \alpha(x)$.) The subcollection $\{U_{\alpha_n}\}_{n \in S}$ is countable, since $S$ is countable. To verify this is a cover: let $x \in X$. By construction, $n(x) \in S$ and $x \in B_{n(x)} \subset U_{\alpha_{n(x)}}$. Hence $x \in U_{\alpha_{n(x)}}$, confirming that $\{U_{\alpha_n}\}_{n \in S}$ covers $X$. Since every open cover admits a countable subcover, $X$ is [Lindelof](/page/Lindel%C3%B6f%20Space).[/step]

[guided]We now select, for each $n \in S$, a single index $\alpha_n \in A$ satisfying $B_n \subset U_{\alpha_n}$. Why does such an $\alpha_n$ exist? Because $n \in S$ means that some $x \in X$ had $n(x) = n$, and for that $x$ we arranged $B_n = B_{n(x)} \subset U_{\alpha(x)}$. So we may take $\alpha_n := \alpha(x)$ for any such $x$. (If multiple points $x$ give the same $n$, we make a single choice; different choices lead to potentially different subcovers, but any choice works.) The resulting family $\{U_{\alpha_n}\}_{n \in S}$ is indexed by the countable set $S$, so it is a countable subcollection of the original cover. We verify it covers $X$: let $x \in X$. Then $n(x) \in S$, and \begin{align*} x \in B_{n(x)} \subset U_{\alpha_{n(x)}}, \end{align*} so $x$ belongs to the subcover. Since $x$ was arbitrary, $\{U_{\alpha_n}\}_{n \in S}$ is a countable subcover. As the open cover $\{U_\alpha\}$ was arbitrary, the space $X$ is [Lindelof](/page/Lindel%C3%B6f%20Space).[/guided]