Is this a non-first-countable space?

Question

I am trying to prove the topological space $(X, \mathcal T)$ below is not a first-countable space. I tried my best but I'm not sure if it is well proved. Whatever there's any improper detail or it's wrong or it is good enough, please tell me.

Thank you very much!

Example

Let $X$ be an uncountable set, and let $\mathcal B$ be the family of all cocountable subsets of $X$. Let $\mathcal T$ be a topology generated by $\mathcal B$. The topological space $(X, \mathcal T)$ is not first-countable.

Proof. Let $\mathcal B_x$ be a countable local base of $x$. Then we have a countable subset of $X$:

$$S = \bigcup_{B_x \in \mathcal B_x} X \setminus B_x = X \setminus \bigcap \mathcal B_x.$$

Now, we are going to show that there exists a required neighbourhood $N$ of $x$ such that for any $B_x \in \mathcal B_x$, $B_x \setminus N$ is not empty.

Let $N \ni x$ be a cocountable proper subset of $X \setminus S$. Clearly, $N$ is a neighbourhood of $x$. Then we have

$$X \setminus S \setminus N = \bigcap \mathcal B_x \setminus N \ne \emptyset.$$

That is, for any $B_x \in \mathcal B_x$, $B_x \setminus N \ne \emptyset$. Thus $(X, \mathcal T)$ is not first-countable.

Answer 1

I’d do the write-up a bit differently: let $x \in X$ arbitrary. We show it does not have a countable local base: striving for a contradiction, assume it has one, say $\{B_n\mid n \in \Bbb N\}$. As $B_n$ is a neighbourhood of $x$ for each $n$ we have a countable set $N_n \subseteq X$ so that

$$x \in X\setminus N_n \subseteq B_n$$

So $N_n, n \in \Bbb N$ is a countable family of at most countable sets, all not containing $x$, so we know that (as $X$ is uncountable) there must be some $y \in X\setminus (\bigcup_{n \in \Bbb N} N_n \cup \{x\})$.

Then $O= X\setminus \{y\}$ is an open subset of $X$ that contains $x$ and as the $\{B_n\mid n \in \Bbb N\}$ is a local base at $x$, for some $m$ we must have $B_m \subseteq O$. But taking complements we then have $\{y\} \subseteq X\setminus B_m \subseteq N_m$ so $y \in N_m$ which is a direct contradiction with how $y$ was chosen. This concludes the proof that $X$ is not first countable at $x$.