Question about Urysohn's metrization theorem When Munkres proves Urysohn's metrization theorem - he does it in steps. He first shows there's countable collection of continuous function with a certain property a every point $x_0$. He starts this in the following way:

Let $\{B_n\}$ be a countable basis for $X$. For each pair $n, m$ of indicies for which $\overline{B}_n \subset B_m$, apply the Urysohn lemma to choose a continuous function with (some properties).

How do we know $\overline{B}_n \subset B_m$? Suppose $X$ is the real line and the basis is all half open unit intervals at every integer. This condition can't be satisfied right?
 A: The space is assumed to be regular, so if $\{B_n\}_{n ∈ ℕ}$ is a basis, then for every $x ∈ B_m$, there exists $n$ such that $x ∈ B_n ⊆ \overline{B_n} ⊆ B_m$. Nobody claims that $\overline{B_n} ⊆ B_m$ for every $n, m$, that's nonsense. The Urysohn lemma is used only for those pairs $n, m$ such that $\overline{B_n} ⊆ B_m$, and regularity asserts that there is enough such pairs. I don't understand the example with unit half-open intervals – they are not open, and they just do not form a basis of the topology of $ℝ$.
A: He doesn't. You could say he defines the set $G=\{(n,m) \in \mathbb{N}: \overline{B_n} \subseteq B_m\}$, the set of "good" pairs. One you've fixed your enumeration of a countable base, there will be some pairs like this, a randomly chosen one won't satisfy it, but there are many pairs that do (regularity will garantuee that, see the end of that proof where regularity is used). At least you know that $G$ is a countable set of pairs and for each of them we pick such a function and we only use those functions.
Munkres said "for each pair where [condition]..." which means he does nothing for the pairs where this is not satisfied, he ignores those.  
