Proving finite complement topology is not A1

In my topology course they show that the first countable axiom isn't valid for the finite complement topology. They state the following:

Suppose X an uncountable set with the following topology,

$$\mathcal{T} = \{ A \subset X | ~X \setminus A ~\text{finite}\}.$$

Take a countable neighbourhood basis $\{B_n | n\in \mathbb{N}\}$ for a neighbourhood filter of $x$, then,

$$X\setminus\{x\} = \cup_n X\setminus B_n$$.

Since the term on the right is an countable union of finite sets and thus countable there follows a contradiction.

Now my question is how do they get the equality?

• $T$ is a $T_1$ topology: The intersection of all the nbhds of a point $p$ is the set $\{p\}.$ – DanielWainfleet Dec 28 '16 at 9:32

It is clear that $X\setminus\{x\}\supset \cup_n X\setminus B_n$, since $x\in B_n$ for all $n$. We need to prove the other inclusion.
Start with a $y\in X\setminus\{x\}$. Now $U=X\setminus\{y\}$ is an open subset of $X$ containing $x$. Hence there is some $B_r$ such that $x\in B_r\subset U$. So, this means that $y\notin B_r$. Thus $y\in X\setminus B_r\subset\cup_n X\setminus B_n$.
We can also do it without the contradiction method: Let $F$ be a countable family of nbhds of $x$. Then $X$ \ $f$ is finite for each $f\in F.$ So $S=\cup_{f\in F}(X$ \ $f)$ is countable. So $S\cup \{x\}$ is countable.
Take any $y\in X$ \ $(S\cup \{x\})$. Then $X$ \ $\{y\}$ is a nbhd of $X.$ But no member of $F$ can be a subset of $X$ \ $\{y\} \;$ (because $\forall f\in F\;(y\in f)\;$). So $F$ is not a nbhd base at $x.$