Compact Space is Countably Compact

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

Then $T$ is countably compact.

Proof
Let $T = \struct {S, \tau}$ be a compact space.

Then by definition every open cover of $S$ has a finite subcover.

So every countable open cover of $S$ has a finite subcover.

Hence by definition $T$ is countably compact.