The problem says
Prove that a countably compact, first countable $T_{2}$ space is regular.
$T_{2}$ space = space Hausdorff
I have I take an open $U$ such that $x \in U$ and $B \subset U$ with $B$ in the neighborhood base of $x.$ Then I take a $W$ such that $V \in W$ with $V$ closed. Now I want to see that $W$ and $U$ are strange, but this happens because the space is Hausdorff
I am right? What else is missing?
Suppose $x \in O$, where $O$ is open, and we want to show that there is some $U$ open such that $$x \in U \subseteq \overline{U} \subseteq O$$
which is an equivalent formulation of being regular. Suppose not, then let $(V_n)_n$ be a decreasing countable local base at $x$ (we can always take finite intersections of initial parts to force the decreasingness), then this failure of regularity says that $\overline{V_n}\setminus O \neq \emptyset$ for all $n$.
In a countably compact space we then conclude that (why?)
$$\exists y \in \bigcap_n (\overline{V_n}\setminus O)$$
Now use Hausdorffness to get a contradiction.