Q:

Will $\{B_n(x)\}_n$ be a local base of point $x$?

If $X$ is a Hausdorff compact space and $\{B_n(x)\}_n$ is a family of open sets of $X$ for each $x\in X$ such that $\cap \{B_n(x)\} =\{x\}$.

Will $\{B_n(x)\}_n$ be a local base of point $x$?

Thanks ahead.

A:

The answer in general (all spaces) is no. Any countable $T_1$ space $X$ has the property that there is a countable family of open sets $B_n(x)$ such that $\cap_n B_n(x) = \{x\}$ for every $x$. Just use $B_n(x) = X\setminus \{y_n\}$ where the $y_n$ enumerate $X\setminus \{x\}$. 
But many countable spaces are not first countable: Arens-Fort space, an ultrafilter space, a countable dense subset of $[0,1]^c$ are some of my favourites. So the above family cannot be a local base for a point where first-countability fails, obviously. All of these spaces are Tychonov, an ultrafilter
space is even (hereditarily) normal.
[added] I later saw you were asking for compact Hausdorff spaces.
There a variant is true: suppose that $O_n(x)$, $n \in \mathbb{N}$ exist such that $\cap_n O_n(x)  =\{x\}$, then $x$ has a countable local base (which need not be the same $O_n$, see Littefield's example. This is known as $\psi\chi(x,X) = \chi(x,X)$ in terms of cardinal invariants. You first make the $O_n$ decreasing in the standard way ,then find open $B_n$ with $x \in B_n \subseteq \overline{B_n} \subseteq O_n$ ($X$ is $T_3$) and use a standard theorem on decreasing sequences of compact sets, to prove that the $B_n$ form a countable local base.

