This is a follow-up to my question here. A subset $A$ of a uniform space is said to be bounded if for each entourage $V$, $A$ is a subset of $V^n[F]$ for some natural number $n$ and some finite set $F$. Just like for metric spaces, a compact subset of a uniform space is always closed and bounded, but the converse need not be true. A uniformity has the Heine-Borel property if a set is compact if and only if it is closed and bounded.
Suppose $X$ is a uniformizable topological space, AKA a completely regular topological space. My question is, is it necessarily the case that the topology on $X$ is induced by some Heine-Borel uniformity?
Now this answer shows that a metrizable topology is induced a metric with the Heine-Borel property if and only if it is locally compact and separable. But that doesn’t answer this question, because even if we take a topology which isn’t locally compact or isn’t separable, it’s possible that such a topology is induced by a non-metrizable uniformity with the Heine-Borel property.
Best Answer
A space $X$ is pseudocompact if any continuous $X\to\mathbb R$ is bounded. There exist completely regular non-compact pseudocompact spaces $X$ such as the long line (there are other examples at π-Base). By pseudocompactness, $X$ is bounded in any continuous pseudometric. By the positive answer to your previous question Is a set bounded in every metric for a uniformity bounded in the uniformity?, noting that the answers also work for pseudometrics in non-metrisable spaces, $X$ is bounded in every uniformity. So $X$ does not admit any Heine-Borel uniformity.