Every Open Cover of $X$ contains a Countable Subcover I am reading this post where George expresses some worries he has concerning a proof in Munkres' Topology book. I myself have a different worry, although I am sure it is related. Here is the relevant passage: 

Let ${B_n}$ be a countable basis and $\mathcal{A}$ an open cover of $X$. For each positive integer $n$ for which it is possible, choose an element $A_n$ of $\mathcal{A}$ containing the basis element $B_n$. The collection $\mathcal{A'}$ of the sets $A_n$ is countable, since it is indexed with a subset $J$ of the positive integers. Furthermore, it covers X: given a point $x \in X$, we can chosse an element $A$ of $\mathcal{A}$ containing $x$. Since $A$ is open, there is a basis element $B_n$ such that $x \in B_n \subset A$. Because $B_n$ lies in an element of $\mathcal{A}$, the index $n$ belong to the set $J$, so $A_n$ is defined; since $A_n$ contains $B_n$, it contains $x$. Thus $\mathcal{A'}$ is a countable subcollection of $\mathcal{A}$ that covers $X$.

I take it that the sentence "For each positive integer $n$ for which it is possible, choose an element $A_n$ of $\mathcal{A}$ containing the basis element $B_n$"  means: Given $B_n \in \mathcal{B}$, where $n \in \Bbb{N}$, if there exists an $A \in \mathcal{A}$ such that $B_n \subseteq A$, then $A_n := A$; and let $\mathcal{A}'$ be the collection of all such $A_n$. I think this is an accurate reformulation (I try to avoid using modal terms in mathematical discourse). My worry is, what if it's never 'possible'; i.e., what if the antecedent is never true?

Look at it this way: Take $x \in X$. As $\mathcal{A}$ is an open cover there is some $A_x \in \mathcal{A}$ that contains $x$. As $\{ B_n: n \in \mathbb{N}\}$ is a base we have that there is some $n_x$ such that $x \in B_{n_x} \subseteq A_x$. So at least for that $n_x$ we have such a member of $\mathcal{A}$ that contains $B_{n_x}$, so being a base forces that this condition will hold often, for many $n$.
The construction just efficiently chooses just $1$ such set for each $n$ for which it is possible, because we want a "small" subcover. We do use the axiom of choice in a strong way (at least countable choice, which most people don't have that much of an issue with).
