Compact Space satisfies Finite Intersection Axiom

Proof
Let every open cover of $S$ have a finite subcover.

Let $\AA$ be any set of closed subsets of $S$ satisfying $\displaystyle \bigcap \AA = \O$.

We define the set:
 * $\VV := \set {S \setminus A : A \in \AA}$

which is trivially an open cover of $S$.

From De Morgan's Laws: Difference with Union:


 * $\displaystyle S \setminus \bigcup \VV = \bigcap \set {S \setminus V : V \in \VV} = \bigcap \set {A : A \in \AA} = \O$

and therefore:
 * $S = \displaystyle \bigcup \VV$

By definition, there exists a finite subcover $\tilde \VV \subseteq \VV$.

We define:
 * $\tilde \AA := \set {S \setminus V : V \in \tilde \VV}$

then $\tilde \AA \subseteq \AA$ by definition of $\VV$.

Because $\tilde \VV$ covers $S$, it follows directly that:


 * $\displaystyle \bigcap \tilde \AA = \bigcap \set {S \setminus V : V \in \tilde \VV} = S \setminus \bigcup \tilde \VV = \O$

Thus, in every set $\AA$ of closed subsets of $S$ satisfying $\displaystyle \bigcap \AA = \O$ exists a finite subset $\tilde \AA$ such that $\displaystyle \bigcap \tilde \AA = \O$.

That is, $S$ satisfies the Finite Intersection Axiom.

The converse works exactly as the previous, but with the roles of the open cover and $\AA$ reversed.

Also see

 * Equivalence of Definitions of Compact Topological Space