Proposition 2

A metrizable space is normal.

Let $X$ be a metrizable space and let $d$ be a metric compatible with the topology of $X$. Let $A, B$ be two disjoint closed subsets of $X$; since the functions $d(x, A)$ and $d(x, B)$ are continuous, the set $U$ (resp. $V$) of points $x$ such that $d(x, A) < d(x, B)$ [resp. $d(x, B) < d(x, A)$] is open; clearly $A \subset U$ and $B \subset V$ and $U \cap V = \varnothing$, hence Axiom $(O'_V)$ is satisfied.