Why does normality imply that a countable base $B$ contains at least one set $U$ whose closure is a subset of another set $V \in B$?

Question

I'm reading Aliprantis and Border's excellent text, Infinite Dimensional Analysis: A Hitchhiker's Guide (PDF available at link, assuming I've done this properly), and I've reached an impasse in the proof of the Urysohn Metrization theorem (3.40, p. 91).

The premises give that $X$ is Hausdorff, and additionally under equivalent statement 3 that $X$ is regular and second countable, so it has a countable base $B$. These conditions also imply by a preceding theorem (2.49, p. 46) that $X$ is normal. For the purposes of the remainder of the proof (which is to show that $X$ can be embedded in the Hilbert cube given regularity and second countability), the authors construct a set $C$ of pairs of nested elements of $B$ such that $C = \{(U,V):\bar{U}\subset V \ \& \ U,V \in B\}$. Then they say, "The normality of $X$ implies that $C$ is nonempty." I've spent considerable time and effort trying to verify this myself, and haven't been able to figure out. It's definitely not a preceding result in this textbook (which is generally very self-contained). And although I've looked around here on Mathoverflow, my topological background might be too weak to see how this is an easy corollary of another theorem.