Pseudocompact Normal Space is Countably Compact

Theorem
Let $T = \left({S, \tau}\right)$ be a normal space.

Then $T$ is pseudocompact $T$ is countably compact.

Proof
Let $T = \left({S, \tau}\right)$ be a normal space.

By Countably Compact Space is Pseudocompact we already have that if $T$ is countably compact then $T$ is pseudocompact.

It remains to prove that if $T$ is pseudocompact then $T$ is countably compact.

$T$ is not countably compact.

By definition of countably compact space, $S$ contains an infinite subset $H = \left\{{x_n}\right\}$ with no $\omega$-accumulation point in $T$.

Since, by definition of normal, $S$ is a $T_1$ space, $H$ is closed and discrete in the subspace topology.

By definition, $T$ is pseudocompact every continuous real-valued function on $S$ is bounded.

Since, by definition of normal, $S$ is a $T_4$ space, the Tietze Extension Theorem guarantees a continuous extension to $S$ of the unbounded continuous mapping $f: H \to \R$ defined by $f \left({x_n}\right) = n$.

This contradicts our hypothesis that $S$ is pseudocompact.

It follows by Proof by Contradiction that $T$ is countably compact.