# Every countably infinite subset of a countably compact space has an $\omega$-cluster point

First the definitions:

• The point $p$ is an $\omega$-cluster point for a subset $A$ of a topological space $X$ if every neighbourhood of $p$ contains infinitely many points of $A$.

• A space is countably compact if every countable open cover has a finite subcover.

Now the question:

How does one prove that every countable infinite subset $A = \{ p_i : i \in \mathbb N \}$ of a countably compact space $X$ has an $\omega$-cluster point?

I know how to prove it if in addition $X$ is assumed to be Hausdorff: by contradiction assuming that there is no $\omega$-cluster point and considering the open cover $\{U_i : i \in \mathbb N \} \cup \{ \mathcal{C}A \}$, where $U_i$ is a neighbourhood of $p_i$ containing at most finitely many points of $A$, and $\mathcal{C}A=X\setminus A$. However $\mathcal{C}A$ is not necessarily open if $X$ is not Hausdorff and I do not know how to modify the proof.

Suppose that $A = \{ p_i : i \in \mathbb N\}$ is a countably infinite subset of a topological space $X$ with no $\omega$-cluster point. For $n \in \mathbb N$ let $F_n := \overline{\{ p_i : i \geq n \}}$. Note that $\{ F_n : n \in \mathbb N \}$ is a decreasing family of nonempty closed subsets of $X$. Note, too, that $\bigcap_{n \in \mathbb N} F_n = \emptyset$:
• Proof. If $x \in \bigcap_{n \in \mathbb N} F_n$, then given any open neighborhood $U$ of $x$ and any $n$, as $x \in F_n = \overline{ \{ p_i : i \geq n \} }$ it must be that $z_i \in U$ for some $i \geq n$. It follows that every open neighborhood of $x$ contains infinitely many $p_i$, which is impossible by assumption.
Therefore $X$ is not countably compact. (By taking the complements of the $F_n$ we get a countable open cover of $X$ with no finite subcover.)