It boils down to the following. (Note: no Hausdorff assumption needed.)
Suppose $X$ is a countable topological space and $A\subseteq X$. If there is a transfinite sequence $(x_\alpha)_{\alpha<\lambda}$ of points of $A$ that converges to a point $p\in X$, then there is an ordinary sequence ($\omega$-indexed) of points of $A$ converging to $p$. Here $\lambda$ is some limit ordinal and saying the transfinite sequence converges to $p$ means: for every nbhd $V$ of $p$, there is some $\beta<\lambda$ such that $x_\alpha\in V$ for all $\alpha>\beta$.
By considering the cofinality of $\lambda$ and taking a corresponding
suitable cofinal subsequence of $(x_\alpha)_{\alpha<\lambda}$, we can assume $\lambda$ is a regular cardinal and then we can further assume it is of smallest (infinite) cardinality satisfying the conditions.
Now, if one of the $x_\alpha=a\in A$ is repeated cofinally many times, then extracting the subsequence of all the elements with value $a$ gives a constant (transfinite) sequence that converges to $p$. (Note that $a$ need not be equal to $p$.) Then, by definition of convergence, every nbhd of $p$ contains $a$ and the ordinary sequence $(y_n)_{n<\omega}$ with each $y_n=a$ also converges to $p$.
Otherwise, for each $a\in A$ the set of indices $I_a=\{\alpha<\lambda: x_\alpha = a\}$ is not cofinal in $\lambda$, that is, $I_a$ has cardinality smaller than $\lambda$. But there are only countably many values $a\in A$. So we get a partition of $\lambda$ into countably many sets, each of cardinality smaller than $\lambda$. If $\lambda$ were uncountable, that would contradict the fact that $\lambda$ is a regular cardinal. So necessarily $\lambda=\omega$ and our original sequence was an ordinary sequence.
This proof also applies to a locally countable space, by first restricting the sequence to a tail contained in a countable nbhd of $p$.
[Added later] From the proof above, all that matters is that $A$ is countable, even if $X$ is not. So basically the result is:
If a point $p\in X$ is the limit of a transfinite sequence (repetitions allowed) of points from a countable set $A\subseteq X$, the point is also the limit of an ordinary sequence of points from $A$.
An attempt at a more handwaving proof would be:
(1) if an element is repeated cofinally many times, do the same as the above. (2) otherwise, for each repeated element in the sequence, take its first appearance and remove all the later copies. The resulting subsequence should also converge to $p$. Why? Because it's cofinal in the original sequence. And why is that? It's not completely obvious, but the argument above justifies it rigorously.
[As mentioned in Steven's comment below, this handwaving attempt does not work.]
(added 11/15/2023)
For more general cardinalities one can use essentially the same argument as above to show for any topological space $X$:
Proposition: Suppose the point $p\in X$ is the limit of a transfinite sequence $(x_\alpha)_{\alpha<\lambda}$ of points from a set $A\subseteq X$ of infinite cardinality $\kappa$. (Here $\lambda$ is some limit ordinal.) Then $p$ is also the limit of a transfinite sequence of points of $A$ indexed by some infinite regular cardinal $\mu$ with $\mu\le\min(\kappa,\lambda)$.
Note: This applies in particular when $A$ is the set of values of the transfinite sequence. Also, if $\mu$ is finite, there is a constant (ordinary) sequence with value in $A$ converging to $p$.
The sequential fan with $\omega$ many spines is a counter-example:
It is defined as follows:
Let $X = \omega \times (\omega+1)$ with the product topology, where both factors have the order topology, i.e., the first factor is discrete, the second one is a convergent sequence.
Identify all points $(n, \omega)$ to a point called $\infty$.
The sequential fan with $\omega$ many spines is the quotient space
$S = S(\omega) = (\omega \times \omega) \cup \{\infty\}$.
Each singleton except $\{\infty\}$ is closed and open in $S$,
hence $S$ is T2, regular and thus even perfectly normal (since it is countable).
Neighborhoods of $\infty$ are of the form
$\{\infty\} \cup \bigcup_{n \in \omega} (\{n\} \times [m_n, \omega))$.
Obviously, each $K_n := ([0, n] \times \omega) \cup \{\infty\}$ is compact (since it is the image of a compact set in $X$).
Moreover, by the above it is easy to see:
- Each compact subset of $S$ is contained in some $K_n$. Hence $S$ is hemicompact, but not locally compact.
- If $\infty \in \overline{B} \setminus B$ for some $B \subset S$, then there is a $n$ such that $(\{n\} \times \omega) \cap B$ is infinite.
Hence $S$ is Fréchet-Urysohn.
Best Answer
Yes, it is well-known that
To be clear about definitions: $X$ is countably compact iff every countable open cover of $X$ has a finite subbcover. I'll use the convenient equivalence that $X$ is countably compact iff every infinite subset $A$ of $X$ has an $\omega$-accumulation point $p \in X$, i.e. every neighbourhood $U$ of $p$ has $U \cap A$ infinite. I wrote the proof of that equivalence (also free of separation axiom assumptions) here, in case this was unknown to you.
And $X$ is sequentially compact iff every sequence $(x_n)_n$ in $X$ has a convergent subsequence.
$X$ is sequential if for all subsets $A$ of $X$: $A$ sequentially closed iff $A$ is closed.
Proof (following the standard reference Engelking, General Topology 2nd ed. Theorem 3.10.31 (due to Franklin (1965), generalising this from metric spaces where it had been shown by Hausdorff in 1914). I'll be extra pedantic in my version of its proof to make sure I make no hidden separation axiom assumptions (Engelking includes Hausdorff in his definition of countable compactness and sequential compactness, so there are subtle instances in his proofs where he might use them, which I want to avoid for maximal generality)
If $X$ is sequentially compact, let $A$ be an infinite subset of $X$. Let $(a_n)_n$ be sequence of points from $A$ so that $a_n \neq a_m$ whenever $n \neq m$ (i.e. find an injection $\Bbb N \to A$). By sequential compactness, there is some $p \in X$ and some subsequence $(a_{n_k})_k$ of $(a_n)_n$ such that $a_{n_k} \to p$ as $k \to \infty$. Then if $U$ is any neighbourhood of $p$, there exists some $N$ so that for all $k \ge N$ we have $a_{n_k} \in U$. It follows that $\{a_{n_k} \mid k \ge N\} \subseteq U \cap A$ and so $U \cap A$ is infinite (by the injectivity). Hence $p$ is an $\omega$-limit point of $A$ and by the mentioned equivalence, $X$ is countably compact. This shows one implication.
Let $X$ be countably compact. Let $(x_n)_n$ be a sequence in $X$. We want to show it has a convergent subsequence.
We can assume WLOG that $n \neq m \to x_n \neq x_m$ (if you believe this read on, if you want my argument for it, reveal spoiler)
Define $A = \{x_n\mid n \in \Bbb N\}$ which is an infinite set. As $X$ is countably compact, $A$ has an $\omega$-accumulation point $p$. It's clear that the $A\setminus \{p\}$ is not closed, as $p$ is in its closure but not in the set. Because $X$ is sequential (finally we use it!), $A\setminus \{p\}$ is not sequentially closed, i.e. there is a sequence $(y_n)_n$ in $A \setminus \{p\}$ and some point $q \notin (A\setminus \{p\})$ so that $y_n \to q$.
By re-ordering this sequence we find a subsequence of $(x_n)_n$ converging to $q$. (details follow as before, skip if you see it already).
QED.