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I have search this question but just found at most some models that it is consistent without any especial example for its existence.

Hussein Mastaneh
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    See if this helps. I haven't looked at it in a while, but recall having written it and was able to find it with a quick google search. (I have a 10:00 AM CST meeting with someone, and thus I'm going out the door right now.) – Dave L. Renfro Sep 12 '19 at 14:36
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    Under ZFC there is no such space: https://mathoverflow.net/questions/38450/compact-hausdorff-spaces-without-isolated-points-in-zf Without the Axiom of Choice things become weird, complicated and you have to be very careful, even with simple definitions. – freakish Sep 12 '19 at 14:44
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    What does "crowded" mean here? – David C. Ullrich Sep 12 '19 at 15:38
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    @DavidC.Ullrich Having no isolated points. It's a pretty standard usage in general topology circles. – Henno Brandsma Sep 12 '19 at 16:40
  • The first part of the 1st answer in the link in the comment from @freakish can also be modified to prove (in ZFC) that a non-empty crowded completely metrizable space has a subspace homeomorphic to the Cantor set. – DanielWainfleet Sep 12 '19 at 18:24
  • Slightly more generally, if X is locally-compact Hausdorff, non-empty, and crowded then (in ZFC) X has an uncountable compact subspace. – DanielWainfleet Sep 12 '19 at 18:27

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If $X$ is (locally) compact Hausdorff it is a Baire space and if it is crowded, it cannot be countable as otherwise $X \setminus \{x\}$, $x \in X$ would be a countable family of open and dense subsets with empty (and not dense) intersection. This uses a mild form of choice (DC ((countable) dependent choice) I think), but that's not a problem IMHO.

If you don't wat to invoke Baire-ness, just mimick its proof: let $f:\mathbb{N} \to X$ be any function and pick (by DC again) for each $n$ some non-empty open set with compact closure (automatic for compact $X$) $O_n$ such that $f(n) \notin \overline{O_n}$ and the $O_n$ are nested. Then $\bigcap_n \overline{O_n} \neq \emptyset$ by compactness and contains a point not in $f[\mathbb{N}]$, implying $X$ cannot be countable. This is in essence Cantor's original proof for the fact that $[0,1]$ is not countable, although he used shrinking balls and Cauchy sequences.

So if you accept a mild form of choice (or full-blown ZFC, as I tend to do) no compact crowded Hausdorff space can be countable, and with some more effort we can even show $|X| \ge 2^{\aleph_0}$, but in ZF, without any choice assumed, I believe countable such spaces can exist IIRC, but I'm no specialist.

Henno Brandsma
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