Proof that metrizable spaces are normal by distance functions

Proof

We separate the two closed sets by comparing the distances to them.

1. Since $X$ is metrizable, choose a metric $d$ on $X$ whose metric topology is the topology of $X$. Let $A, B \subseteq X$ be disjoint closed sets.
2. If $A = \varnothing$, take $U = \varnothing$ and $V = X$. If $B = \varnothing$, take $U = X$ and $V = \varnothing$. From now on let $A$ and $B$ be nonempty.
3. For nonempty $C \subseteq X$, write $d(x, C)$ for the distance from x to C. The metric axioms give $d \ge 0$, so $d(x, C) \ge 0$.
4. Claim: $d(y, C) \le d(y, x) + d(x, C)$ for all $x, y \in X$. For each $c \in C$, the triangle inequality gives $d(y, C) \le d(y, c) \le d(y, x) + d(x, c)$.
5. So $d(y, C) - d(y, x)$ is a lower bound of $\set{d(x, c) : c \in C}$. By the definition of the infimum, it is at most $d(x, C)$. This proves the claim.
6. Claim: if $C$ is closed and $x \notin C$, then $d(x, C) > 0$. The set $X \setminus C$ is open and contains $x$, so some open ball $B(x, r)$ with $r > 0$ misses $C$.
7. Then $d(x, c) \ge r$ for each $c \in C$, so $d(x, C) \ge r > 0$.
8. Let $U = \set{x : d(x, A) < d(x, B)}$ and $V = \set{x : d(x, B) < d(x, A)}$. No point satisfies both inequalities, so $U \cap V = \varnothing$.
9. Let $a \in A$. Then $d(a, A) = 0$, since $d(a, a) = 0$. Since $a \notin B$ and $B$ is closed, step 6 gives $d(a, B) > 0$. So $a \in U$, and $A \subseteq U$.
10. In the same way, $B \subseteq V$.
11. We show that $U$ is open. Let $x \in U$ and $\eta = d(x, B) - d(x, A) > 0$. Let $y \in B(x, \eta / 2)$.
12. By step 4, $d(y, A) \le d(x, A) + d(x, y)$. Also by step 4, $d(x, B) \le d(x, y) + d(y, B)$, so $d(y, B) \ge d(x, B) - d(x, y)$.
13. Subtracting, $d(y, B) - d(y, A) \ge \eta - 2 d(x, y) > 0$. So $y \in U$.
14. Thus $B(x, \eta / 2) \subseteq U$ for each $x \in U$, and $U$ is open in the metric topology. In the same way, $V$ is open.
15. So $U$ and $V$ are disjoint open sets with $A \subseteq U$ and $B \subseteq V$. Hence $X$ is normal.