Countably Compact Space satisfies Countable Finite Intersection Axiom

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

Let $\AA$ be any set of closed sets of $T$ satisfying $\bigcap \AA = \O$.

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

which is clearly an open cover of $S$.

From De Morgan's Laws: Difference with Union:


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

and therefore $S = \bigcup \VV$.

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:


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

Thus, in every countable set $\AA$ of closed sets of $T$ satisfying $\ds \bigcap \AA = \O$ exists a finite subset $\tilde \AA$ such that $\ds \bigcap \tilde \AA = \O$.

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

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