Intersection of Closed Set with Compact Subspace is Compact/Proof 2

Proof
Let $\family {U_\alpha}$ be an open cover of $H \cap K$:
 * $\ds H \cap K \subseteq \bigcup_\alpha U_\alpha$

Then:
 * $\ds K \subseteq \bigcup_\alpha U_\alpha \cup \paren {S \setminus H}$

Since $H$ is closed in $T$, $\paren {S \setminus H}$ is open in $T$.

Hence $\family {U_\alpha} \cup S \setminus H$ is an open cover of $K$.

We have that $K$ is compact in $T$.

It follows by definition that a finite subcover:
 * $\set {U_{\alpha_1}, U_{\alpha_2}, \ldots, U_{\alpha_n}, S \setminus H}$

of $K$ exists.

Thus:
 * $H \cap K \subseteq \set {U_{\alpha_1}, U_{\alpha_2}, \ldots, U_{\alpha_n} }$

and $H \cap K$ is compact in $T$.