Disjoint Compact Sets in Hausdorff Space have Disjoint Neighborhoods

Theorem
Let $T = \struct {S, \tau}$ be a Hausdorff space.

Let $V_1$ and $V_2$ be compact sets in $T$.

Then $V_1$ and $V_2$ have disjoint neighborhoods.

Proof
Let $\FF$ be the set of all ordered pairs $\tuple {Z, W}$ such that:
 * $Z, W \in \tau$
 * $V_1 \subseteq Z$
 * $Z \cap W = \O$

By the lemma, $\Img \FF$ covers $V_2$.

By the definition of compact space, there exists a finite subset $K$ of $\Img \FF$ which also covers $V_2$.

By the definition of topology, $\ds \bigcup K$ is open.

By the Principle of Finite Choice, there exists a bijection $\GG \subseteq \FF$ such that $\Img \GG = K$.

Then $\GG$, and hence its preimage, will be finite.

Let $\ds J = \bigcap \Preimg \GG$.

By Intersection is Largest Subset, $V_1 \subseteq J$.

By the definition of a topology, $J$ is open.

Then $\ds \bigcup K$ and $J$ are disjoint open sets such that $\ds V_2 \subseteq \bigcup K$ and $V_1 \subseteq J$.