Pseudocompact Normal Space is Countably Compact

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

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

Proof
Let $T = \left({X, \tau}\right)$ 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.

Suppose this were not the case.

Then $X$ would contain an infinite subset $S = \left\{{x_n}\right\}$ with no $\omega$-accumulation point.

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

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

So $X$ could not have been pseudocompact after all.