Metrizable spaces are normal

Statement

Let $X$ be a metrizable space. Then $X$ is normal.

Remarks

Proof: choose a metric $d$ for the topology. For disjoint closed sets $A$ and $B$, let $U = \{x : d(x, A) < d(x, B)\}$ and $V = \{x : d(x, B) < d(x, A)\}$. Here $d(x, A)$ is the distance to a set. The sets $U$ and $V$ are open, because $x \mapsto d(x, A)$ is continuous. They are disjoint, and $A \subseteq U$ because $d(x, A) = 0 < d(x, B)$ for $x \in A$. Single points are closed in a metric space. Thus each metrizable space is normal and T1, and also Hausdorff.

Explanations

L1 human

In a metric space, you separate two disjoint closed sets by the points that are nearer to one set than to the other.