Metric Space is Perfectly T4

Theorem
Let $M = \left({A, d}\right)$ be a metric space.

Then $M$ is a perfectly $T_4$ space.

Proof
We have that a metric space is $T_4$.

We also have that every closed set in a metric space is a $G_\delta$ set.

Hence the result, by definition of a perfectly $T_4$ space.