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:
 * $\ds \OO = \set {V \in \tau: \exists U \in \tau: x \in U \subseteq V^\complement}$

By Empty Intersection iff Subset of Complement:
 * $U \subseteq V^\complement \iff U \cap V = \O$

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

By the definition of a compact subspace, there exists a finite subcover $\FF$ of $\OO$ for $C$.

By the Principle of Finite Choice, there exists an $\FF$-indexed family $\family {U_V}_{V \mathop \in \FF}$ of elements of $\tau$ such that:
 * $\forall V \in \FF: x \in U_V \subseteq V^\complement$

Define:
 * $\ds U = \bigcap_{V \mathop \in \FF} U_V$

By General Intersection Property of Topological Space:
 * $U \in \tau$

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:
 * $C^\complement \in \tau$

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