Compact Subspace of Hausdorff Space is Closed/Proof 2

Proof
For a subset $S \subseteq A$, let $S^{\complement}$ denote the relative complement of $S$ in $A$.

Consider an arbitrary point $x \in C^{\complement}$.

Define the set:
 * $\displaystyle \mathcal O = \left\{{V \in \tau: \exists U \in \tau: x \in U \subseteq V^{\complement}}\right\}$

By Empty Intersection iff Subset of Complement, we have that:
 * $U \subseteq V^{\complement} \iff U \cap V = \varnothing$

Hence, by the definition of a Hausdorff space, it follows that $\mathcal O$ is an open cover for $C$.

By the definition of a compact subspace, there exists a finite subcover $\mathcal F$ of $\mathcal O$ for $C$.

By the Principle of Finite Choice, there exists an $\mathcal F$-indexed family $\left\langle{U_V}\right\rangle_{V \in \mathcal F}$ of elements of $\tau$ such that:
 * $\forall V \in \mathcal F: x \in U_V \subseteq V^{\complement}$

Define:
 * $\displaystyle U = \bigcap_{V \mathop \in \mathcal F} U_V$

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

Clearly, $x \in U$.

We have that:

From Subset Relation is Transitive, we have that $U \subseteq C^{\complement}$.

Hence $C^{\complement}$ is a neighborhood of $x$.

From Set is Open iff Neighborhood of all its Points, we have that $C^{\complement} \in \tau$.

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