Intersection of Closed Set with Compact Subspace is Compact

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.

Let $H \subseteq S$ be closed in $T$.

Let $K \subseteq S$ be compact in $T$.

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

Proof
Let $\left\langle{U_\alpha}\right\rangle$ be an open cover of $H \cap K$:
 * $\displaystyle H \cap K \subseteq \bigcup_\alpha U_\alpha$

Then:
 * $\displaystyle K \subseteq \bigcup_\alpha U_\alpha \cup \left({S \setminus H}\right)$

Since $H$ is closed in $T$, $\left({S \setminus H}\right)$ is open in $T$.

Hence $\left\langle{U_\alpha}\right\rangle \cup S \setminus H$ is an open cover of $H$.

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

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

of $H$ exists.

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

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