Pseudocompactness does not imply compactness

A favorite example (and counterexample) to may things is the first uncountable ordinal $\omega_1$ in its order topology: $[0,\omega_1)$. It is pseudo-compact but not compact.

$\pi$-Base, a searchable version of Steen and Seebach's Counterexamples in Topology, gives the following examples of pseudocompact spaces that are not compact. You can view the search result to learn more about any of these spaces.

$[0,\Omega) \times I^I$

An Altered Long Line

Countable Complement Topology

Countable Particular Point Topology

Deleted Tychonoff Plank

Divisor Topology

Double Pointed Countable Complement Topology

Gustin’s Sequence Space

Hewitt's Condensed Corkscrew

Interlocking Interval Topology

Irrational Slope Topology

Minimal Hausdorff Topology

Nested Interval Topology

Novak Space

Open Uncountable Ordinal Space $[0, \Omega)$

Prime Integer Topology

Relatively Prime Integer Topology

Right Order Topology on $\mathbb{R}$

Roy's Lattice Space

Strong Ultrafilter Topology

The Long Line

Tychonoff Corkscrew

Uncountable Particular Point Topology

I couldn't think of an obvious counterexample, so I looked in Wikipedia and it suggested the particular point topology on an infinite set.

$\def\p{{\bf x}}$In the particular point topology, we have a distinguished point, $\p\in X$, and the topology is that a set is open if and only if it is either empty, or includes $\p$.

Let $S=\langle X,{\mathfrak I}\rangle$ be an infinite particular-point space with distinguished point $\p$. It is clear that $S$ is not compact: the open cover consisting of $\{p, \p\}$ for each $p\in X$ other than $\p$ is an infinite open cover of $S$ with no proper, and therefore no finite subcover.

$\def\R{{\Bbb R}}$However, the space is pseudocompact. Let $f:S\to\R$ be a continuous function. Then $f^{-1}[\Bbb R\setminus \{f(\bf x)\}]$ is an open set not containing $\bf x$, so it must be empty, hence $f$ is constant.