Countably Compact Space is Countably Paracompact

Theorem
Let $T = \left({X, \tau}\right)$ be a countably compact space.

Then $T$ is countably paracompact.

Proof
From the definition, $T$ is countably compact every countable open cover of $X$ has a finite subcover.

From Subcover is Refinement of Cover, it follows that every countable open cover of $X$ has an open refinement which is locally finite.

This is precisely the definition of countably paracompact.