Prove that if $X$ is compact space, $x\in X, \{x\}\text{ is a }G_{\delta}$ set, it follows that $\chi(x,X)\leq\aleph_{0}$ Prove that if $X$ is compact space, $x\in X, \{x\}\text{ is a }G_{\delta}$ set, it follows that $\chi(x,X)\leq\aleph_{0}$, where
$\chi(x,X):=min\{|B(x)| : B(x)\text{ is local base at x}\}$ and
$x-G_{\delta}$ set if $x=\cap_{i=1}^{\infty}U_{i} $ for some $U_i$ open subsets of X.
I have no idea how to approach this problem. Compact space means that every open cover has a finite subcover, or equivalently, every family of closed sets with the finite intersection property has nonempty intersection, but I don't quite see how it helps here.
 A: We need $X$ to be Hausdorff, and this then also implies $X$ is regular (even normal, which I don't need).
Write $\{x\} = \bigcap_{n=1}^\infty U_n$.
Now using regularity we find for each $n$ an open subset $O_n$ such that $x \in O_n \subseteq \overline{O_n} \subseteq U_n$, and by taking finite intersections we can ensure that moreover the $O_{n+1} \subseteq O_n$ for all $n$ as well. Note that we still have that $\bigcap_n \overline{O_n} = \{x\}$.
The $O_n$ now form a local base at $x$. 
Proof: let $O$ be an open subset of $X$ that contains $x$, and we want to find some $O_n$ such that $x \in O_n \subseteq O$. Suppose, for a contradiction, that no such such $n$ exists, so that $O_n \setminus O$ is always non-empty.
But then $C_n = \overline{O_n}\setminus O$ is a decreasing sequence of closed non-empty sets in $X$, so by a standard fact by compactness $\cap_n C_n \neq \emptyset$ (the $C_n$ have the finite intersection property; or consider the open cover of the complements if you assume an empty intersection). But this is a contradiction as $\bigcap_n C_n = (\bigcap_n \overline{O_n}) \setminus O = \{x\} \setminus O = \emptyset$. So we must have that for some $n$ (and thus also all larger ones) $O_n \subseteq O$, and we have a local base.
