One-point sets are G$_\delta$ in first-countable $T_1$ spaces

Question

Can someone please verify my proof or offer suggestions for improvement? I am aware that there is a similar question elsewhere. I only need help with my proof in particular.

Show that in a first-countable $T_1$ space, every one-point set is a $G_\delta$ set.

Let $X$ be a first-countable $T_1$ space. Let $x \in X$. Then, there exists a countable basis $\{B_n\}$ at $x$. Pick $y \neq x$. Since $X$ is $T_1$, $X -\{y\}$ is a neighborhood of $x$. So, there exists a $B_n$ such that $B_n \subseteq X - \{y\}$. But then, $y \notin B_n$. So, $y \notin \displaystyle{\bigcap_{i = 1}^\infty B_i}$. Clearly, $x \in \displaystyle{\bigcap_{i = 1}^\infty B_i}$. So, $\displaystyle{\bigcap_{i = 1}^\infty B_i} = \{x\}$.

Answer 1

Your proof is correct. I think that my first sentence is full answer to your question, however validation says it must be at least 30 characters long, so I'm writing this. :-)

Answer 2

A more complete proof may be like this.

A first countable space $X$ has a countable local basis at every point. Let $x \in X$ and let $\mathcal{B}_x = \{B_n : n\in \mathbb{N}\}$ denote the set of local basis at $x$. Pick $y \neq x$. Since, $X$ is $T_1$, the set $X \setminus \{y\}$ is a neighborhood of $x$. Again, by property of local basis, there exists a $B_n$ such that $B_n \subseteq X \setminus \{y\}$. Call this $B_n$ to be $B_{n_{(y)}}$; therefore, $y \notin B_n$ and hence, $y \notin B_{n_{(y)}}$. Since, for every $y$, $y \notin B_{n_{(y)}}$, but $x \in B_{n_{y}}$ we must have $x \in \underset{y \ne x}{\bigcap} B_{n_{(y)}}$. Let's assume for the sake of contradiction that $\exists \ z $ such that $z \in \underset{y \ne x}{\bigcap} B_{n_{(y)}}$, then the only possible choice for $z$ is $x$ since the intersection is overall $y\ne x$. Thus, $\underset{y}{\bigcap} B_{n_{(y)}} = \{x\}$ and $\{x\}$ is $G_{\delta}$ as required.