Question about compact space

Question

Let $(X, \tau) $ a compact Hausdorff topological space such that for all $x \in X$ exists $U_x \subset \tau$ countable such that $\bigcap U_x =\{ x \} $. How can I prove that $(X, \tau) $ is a first-countable space?

Answer 1

Define $\mathcal{B}$ to be all finite intersections of members of $U_x$ so $$\mathcal{B} = \{\cap \mathcal{F}: \mathcal{F} \subseteq \mathcal{U_x} \text{ finite }\}$$ and note that this is still a countable family of open neighbourhoods of $x$, and moreover it's closed under finite intersections.

Also note that Hausdorfness implies that

$$\cap \{\overline{O} \in U_x\} = \cap_n \{\overline{B}: B \in \mathcal{B}\} = \{x\}$$

(any $q \neq x$ has a disjoint neighbourhood with some neighbourhood from $U_x \subseteq \mathcal{B}$ and so does not lie in either intersection.) Now let's see we have a local base at $x$:

Suppose that $x \in O$, where $O$ is open. We want to show that there is some member $B \in \mathcal{B}$ such that $x \in B \subseteq O$.

Suppose that this were not the case. Then $\{X\setminus \overline{B}: B \in \mathcal{B}\} \cup\{O\}$ is an open cover of $X$ (if $p$ is in $X$ and $p \notin X\setminus \overline{B}$ for all $B \in \mathcal{B}$, then $p =x$ by the intersection property and so $p \in O$). So finitely many of them cover $X$ by compactness, say the sets $O$ and $X\setminus \overline{B_{n_1}}, \ldots, X\setminus \overline{B_{n_k}}$, for $B_{n_i} \in \mathcal{B}, i=1\ldots k$ But that these sets are a cover of $X$ implies that $\mathcal{B} \ni \cap_{i=1}^k B_{n_i} \subseteq O$ (if $p \in X$ is in all of the $B_{n_i}$, it's in none of the $X\setminus \overline{B_{n_i}}$, so it must be in $O$). This contradiction shows the claim.