Metric Space is T4/Proof 1

Proof
Let $H$ and $K$ be disjoint closed sets of $M$.

Let $g: A \to \R$ be defined as:
 * $g = f_K - f_H$

where:
 * $\forall x \in A: \map {f_K} x = \map d {x, K}$
 * $\forall x \in A: \map {f_H} x = \map d {x, H}$

where $\map d {x, K}$, $\map d {x, H}$ denotes the distance from $x$ to $K$ and from $x$ to $H$ respectively.

From Distance from Point to Subset is Continuous Function, both $\map {f_K} x$ and $\map {f_H} x$ are continuous mappings.

Hence $g$ itself is a continuous mapping.

Consider the disjoint open intervals $\openint 0 \to$ and $\openint \gets 0$.

Because $g$ is continuous, it follows that $g^{-1} \sqbrk {\openint 0 \to}$ and $g^{-1} \sqbrk {\openint \gets 0}$ are open in $M$ and disjoint.

Let $x \in H$.

Then:
 * $\map {f_H} x = 0$

Because $H$ and $K$ are disjoint:
 * $H \cap K = \O$

and so:
 * $x \notin K$

Because $K$ is closed sets in $M$, it follows from Point at Distance Zero from Closed Set is Element that:
 * $\map {f_K} x > 0$

It follows that:
 * $\map g x > 0$

that is:
 * $H \subseteq g^{-1} \sqbrk {\openint 0 \to}$

Similarly:
 * $K \subseteq g^{-1} \sqbrk {\openint \gets 0}$

The result follows.