If $\displaystyle\bigcap_{n\in\mathbb{N}} U_n=\{x\}$, then $\{U_1,U_2,...\}$ is a local base at $x$. 
Let $X$ be a countably compact space, $x\in X$, and $U_1,U_2,...$ open sets such that $\displaystyle\bigcap_{n\in\mathbb{N}} U_n=\{x\}$. Then $\{U_1,U_2,...\}$ is a local base at $x$.

We take an open set $A\subseteq X$ with $x\in A$. So $\displaystyle\bigcap_{n\in\mathbb{N}} U_n\subseteq A$. I want to prove there's some $n\in\mathbb{N}$ such that $x\in U_n\subseteq A.$
I don't really know what we could do here.
Countably compact is every open countable cover has a finite subcover. I also have proved that it's equivalently to each sequence has an accumulation point. And closed subspaces of $X$ are also countably compact.
Could someone give a hint?
 A: This is definitely false as stated. Let $\tau$ be the cofinite topology on $\Bbb N$; $\langle\Bbb N,\tau\rangle$ is compact. For $n\in\Bbb Z^+$ let $U_n=\Bbb N\setminus\{n\}$; the sets $U_n$ are open, and $\bigcap_{n\in\Bbb Z^+}U_n=\{0\}$, but $\Bbb N\setminus\{1,2\}$ is an open nbhd of $0$ that does not contain any $U_n$. 
You can avoid this kind of trivial counterexample by requiring in the statement of the result that $U_n\supseteq U_{n+1}$ for each $n\in\Bbb N$. However, even this isn’t enough. J.E. Vaughan, Countably compact, locally countable $T_2$-spaces [PDF], Proc. Amer. Math. Soc. $80$ ($1980$), pp. $147$-$153$, constructs a countably compact Hausdorff space $X$ of cardinality $2^{\mathfrak{c}}$ in which every singleton is the intersection of a countable family of open nbhds, but there are no non-trivial convergent sequences. Clearly $X$ is not discrete, so it must have at least one limit point $x$. If there were a countable local base at $x$, we could construct a sequence in $X\setminus\{x\}$ converging to $x$; since $X$ has no non-trivial convergent sequences, however, this is impossible, and therefore $x$ cannot have a countable local base.
