Metrizable implies monotonically normal

Statement
Any metrizable space is monotonically normal. In fact, we can construct an explicit monotone normality operator using the metric.

Construction of the operator
Suppose $$(X,d)$$ is a metric space. We construct a monotone normality operator on $$X$$ as follows. For any two closed subsets $$A$$ and $$B$$ of $$X$$:


 * For every $$a \in A$$, let $$U_a$$ be the open ball about $$a$$ of radius $$d(a,B)/2$$. Here, $$d(a,B)$$ is the infimum of the distances $$d(a,b)$$ for $$b \in B$$. Note that since $$B$$ is closed and $$a \notin B$$, $$d(a,B) > 0$$.
 * Analogously, for every $$b \in B$$, define $$V_b$$ to be the open ball about $$b$$ of radius $$d(A,b)/2$$.

Then, the sets:

$$U = \bigcup_{a \in A} U_a, \qquad V = \bigcup_{b \in B} V_b$$

are the required disjoint open subsets.

Proof of disjointness
The disjointness of $$U$$ and $$V$$ follows from the triangle inequality. It suffices to prove that for every $$a \in A$$ and $$b \in B$$, $$U_a \cap V_b = \emptyset$$. To see this, note:

$$d(a,b) \ge d(a,B), \qquad d(a,b) \ge d(A,b)$$

Thus:

$$d(a,b) \ge d(a,B)/2 + d(A,b)/2$$

This, along with the triangle inequality yields that the balls $$U_a$$ and $$V_b$$ cannot intersect.

Proof of monotonicity
We need to show that if $$A' \subseteq A$$ and $$B' \supseteq B$$, then the open set about $$A$$ becomes smaller, and the open set about $$B$$ becomes larger. This is clear from the definitions, because $$A'$$ has fewer balls, and the radii of the balls also become smaller. Similarly $$B'$$ has more balls, and the radii of the balls also become larger.