Pseudocompactness does not imply compactness It is well known that compactness implies pseudocompactness; this follows from the Heine–Borel theorem. I know that the converse does not hold, but what is a counterexample?
(A pseudocompact space is a topological space $S = \langle X,{\mathfrak I}\rangle$ such that every continuous function $f:S\to\Bbb R$ has bounded range.)
 A: 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.
A: $\pi$-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
A: An example that does not depend on countable compactness is Mrówka’s space $\Psi$. Subsets of $\omega$ are said to be almost disjoint if their intersection is finite. Let $\mathscr{A}$ be a maximal almost disjoint family of subsets of $\omega$, and let $\Psi=\omega\cup\mathscr{A}$. Points of $\omega$ are isolated. Basic open nbhds of $A\in\mathscr{A}$ are sets of the form $\{A\}\cup(A\setminus F)$, where $F$ is any finite subset of $A$. $\Psi$ is not even countably compact, since $\mathscr{A}$ is an infinite (indeed uncountable) closed, discrete set in $\Psi$. (In fact it’s not hard to ensure that $|\mathscr{A}|=2^\omega$.)
To see that $\Psi$ is pseudocompact, suppose that $f:\Psi\to\Bbb R$ is continuous. Since $\omega$ is dense in $\Psi$, it suffices to show that $f[\omega]$ is bounded. If not, we can choose $S=\{n_k:k\in\omega\}\subseteq\omega$ such that $f(n_{k+1})\ge f(n_k)+1$ for each $k\in\omega$. The maximality of $\mathscr{A}$ ensures that there is an $A\in\mathscr{A}$ such that $A\cap S$ is infinite. Let $A_0=\{k\in\omega:n_k\in A\cap S\}$. Then $\langle f(n_k):k\in A_0\rangle\to f(A)$, which contradicts the choice of $S$.
$\Psi$ clearly is $T_2$ and has a clopen base, so it’s Tikhonov. It’s not normal, however, since in $T_4$ spaces pseudocompactness is equivalent to countable compactness.
Added: This example is somewhat akin to what Steen & Seebach call the strong ultrafilter topology.
A: Note that sequential compactness also implies pseudocompactness, so any sequentially compact space which is not compact will work as well (the particular point topology is not sequentially compact, either, so it's different kind).
For example, the Corson space $\Sigma([0,1]^\kappa)$ of sequences of length $\kappa$ with countable support is not compact for uncountable $\kappa$ (which is easy to see), but is sequentially compact (which is a bit harder to see). It is also completely regular Hausdorff, which makes it sort of a "stronger" example than particular point topology. $\Sigma(2^\kappa)$ should work fine for this, too.
A: Some exotic examples of pseudocompact and non-compact spaces are constructed in my paper “Pseudocompact paratopological groups that are topological”: 
Example 1 (p.6). A $T_1$ space having each power countably pracompact (and, hence pseudocompact). (See p.1 for the definition of a countably pracompact space).
Example 2 (p.6). [Under $MA_{countable}$] A functionally Hausdorff countably compact space $X$. (See p.6 for axiomatic assumptions which I use for the example construction).
Example 3 (p.8) A functionally Hausdorff second countable space having each power countably pracompact (and, hence pseudocompact).
Example 5 (p. 14) A $T_0$ sequentially compact, not totally countably compact space and a $T_1$ pseudocompact, not countably pracompact space. (See p.1 for the definition of a totally countably compact space).
A: 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.
A: This nice paper presents all kinds of relations between different compactness notions.
http://www.cs.cmu.edu/~yaoliang/mynotes/compact.pdf
E.g., if a topological space $K$ is compact or sequentially compact, then it is countably compact. If $K$ is countably compact, then it is pseudo-compact.
So pick any countably or sequentally compact space that is not compact.
Some are given here: Limit point Compactness does not imply compactness counter-example
Or pick a limit-point compact space which is T1 or N1 (hence pseudo-compact) but not compact.
Some are given here: Prove or disprove : limit point compact hausdorff space imply compact space?
N.B. A continuous image of any compact space is compact. A compact subset of ${\mathbb R}$ is bounded. Therefore, compactness implies pseudo-compactness.
