Completely Hausdorff Property is Hereditary

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space which is a $T_{2 \frac 1 2}$ (completely Hausdorff) space.

Let $T_H = \left({H, \tau_H}\right)$, where $\varnothing \subset H \subseteq S$, be a subspace of $T$.

Then $T_H$ is a $T_{2 \frac 1 2}$ (completely Hausdorff) space.

Proof
Let $T = \left({S, \tau}\right)$ be a $T_{2 \frac 1 2}$ (completely Hausdorff) space.

Then:
 * $\forall x, y \in S, x \ne y: \exists U, V \in \tau: x \in U, y \in V: U^- \cap V^- = \varnothing$

That is, for any two distinct points $x, y \in S$ there exist open sets $U, V \in \tau$ containing $x$ and $y$ respectively whose closures are disjoint.

We have that the set $\tau_H$ is defined as:
 * $\tau_H := \left\{{U \cap H: U \in \tau}\right\}$

We can assume that $\varnothing \subset H \subset S$, that is, $H \ne \varnothing$ and $H \ne S$, otherwise the result is trivial.

Let $x, y \in H$ such that $x \ne y$.

Then as $x, y \in S$ we have that:
 * $\exists U, V \in \tau: x \in U, y \in V, U^- \cap V^- = \varnothing$

As $x, y \in H$ we have that:
 * $x \in U \cap H, y \in V \cap H: \left({U^- \cap H}\right) \cap \left({V^- \cap H}\right) = \varnothing$

Thus:
 * $x \in U \cap H, y \in V \cap H: \left({U \cap H}\right)^- \cap \left({V \cap H}\right)^- = \varnothing$

and so the $T_{2 \frac 1 2}$ (completely Hausdorff) axiom is satisfied in $H$.