Compact Subspace of Hausdorff Space is Closed

Theorem
Let $\left({H, \vartheta}\right)$ be a Hausdorff space.

Let $C$ be a compact subspace of $H$.

Then $C$ is closed in $H$.

Proof
For any element $x \in H \setminus C$ of the relative complement of $C$ in $H$, define:
 * $\displaystyle \mathcal C \left({x}\right) = \left\{{V \in \vartheta: \exists U \in \vartheta: x \in U, \, U \cap V = \varnothing}\right\}$

By the definition of a Hausdorff space, it follows that $\mathcal C \left({x}\right)$ is an open cover for $C$.

By the definition of a compact space, there exists a finite subcover $\mathcal F$ of $\mathcal C \left({x}\right)$ for $C$.

Since $C$ is non-empty, it follows by Union of Empty Set that $\mathcal F$ is non-empty.

For all $V \in \mathcal F$, define:
 * $\mathcal A \left({V}\right) = \left\{{U \in \vartheta: x \in U, \, U \cap V = \varnothing}\right\}$

By the definition of $\mathcal C \left({x}\right)$, it follows that $\mathcal A \left({V}\right)$ is non-empty.

By Finite Cartesian Product of Non-Empty Sets is Non-Empty, there exists a family $\left\langle{U \left({V}\right)}\right\rangle_{V \in \mathcal F}$ such that:
 * $\forall V \in \mathcal F: U \left({V}\right) \in \vartheta, \, x \in U \left({V}\right), \, U \left({V}\right) \cap V = \varnothing$

Define:
 * $\displaystyle U = \bigcap_{V \mathop \in \mathcal F} U \left({V}\right)$

By General Intersection Property of Topological Space, it follows that $U \in \vartheta$.

Clearly, $x \in U$.

Therefore, $U \subseteq H \setminus C$.

Hence, $x \in \left({H \setminus C}\right)^{\circ}$, where $^{\circ}$ denotes interior.

That is, $H \setminus C \subseteq \left({H \setminus C}\right)^{\circ}$.

By Set Interior is Largest Open Set, $\left({H \setminus C}\right)^{\circ} \subseteq H \setminus C$.

Therefore, by Equality of Sets, $H \setminus C = \left({H \setminus C}\right)^{\circ}$.

Hence, by Interior of Open Set, $H \setminus C \in \vartheta$.

That is, $C$ is closed in $H$.