All normal pseudocompacts are weakly countably compact (that is, infinite subsets cannot be closed and discrete).
Can "normal" be weakened to "pseudonormal"? A space is pseudonormal provided for every countable closed set $H$ and open $U\supseteq H$, there is an open set $V$ with $H\subseteq V\subseteq cl(V)\subseteq U$.
Say that $X$ is completely regular for closed points if for any closed $x\in X$ and disjoint closed $F\subseteq X$ there is a continuous function $f:X\to [0, 1]$ such that $f(x) = 0$ and $f\equiv 1$ on $F$.
Say that $X$ is regular for closed points if for any closed $x\in X$ and disjoint closed $F\subseteq X$ there are disjoint open $U, V$ such that $x\in U$ and $F\subseteq V$.
Note that both pseudonormal and completely regular for closed points imply regular for closed points.
Also both normal and completely regular implies completely regular for closed points.
Lemma. Let $X$ be regular for closed points and $D\subseteq X$ be countable closed and discrete. Then there are disjoint open $U_x, x\in D$ with $x\in U_x$.
Proof: Proceed by induction. $\square$
Proposition. Suppose that $X$ is pseudonormal, completely regular for closed points, and $D\subseteq X$ is a closed countably infinite discrete set. Then $D$ is $C$-embedded in $X$.
Proof:
Let $D = \{x_1, x_2, ...\}$. For each $n$ take open $U_n$ with $x_n\in U_n$ and all $U_n$ disjoint. Since $X$ is pseudonormal, we can take open $U$ such that $D\subseteq U\subseteq \overline{U}\subseteq \bigcup_n U_n$. Let $g:D\to\mathbb{R}$. Let $f_n:X\to \mathbb{R}$ be continuous such that $f_n(x_n) = g(x_n)$ and $f_n\equiv 0$ for $(U_n\cap U)^c$. Let $f\equiv f_n$ on $U_n$ and $f\equiv 0$ on $U^c$, then $f$ is well-defined. Moreover, $f \restriction_{U_n}$ and $f\restriction_{\overline{U}^c}$ are continuous. From gluing lemma, $f$ is continuous. $\square$
Theorem. Let $X$ be a pseudocompact, pseudonormal and completely regular for closed points. Then $X$ is weakly countably compact.
Proof:
If $X$ is not weakly completely compact, let $D\subseteq X$ be closed countably infinite and discrete. From the lemma, $D$ is $C$-embedded in $X$. But by picking an unbounded function on $D$ and extending it to $X$, this contradicts pseudocompactness. $\square$
Corollary. Let $X$ be either a completely regular pseudonormal space, or a normal space. If $X$ is pseudocompact, then $X$ is weakly countably compact.