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?

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 $y\in X\setminus B_r\subset\cup_n X\setminus B_n$.
