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.