Countably Compact Space is Pseudocompact

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

Then $T$ is a pseudocompact space.

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

Then every countable open cover of $X$ has a finite subcover.

By definition:
 * $T$ is pseudocompact every continuous real-valued function on $X$ is bounded.

Let $f: X \to \R$ such that $f$ be continuous.

Consider the sets:
 * $S_n = \left\{{x \in X: \left \vert {f \left({x}\right)}\right \vert < n}\right\}$

$S_n$ forms a countable cover of $X$ whose finite subcover yields a bound for the absolute value of $f$.

Hence the result.