Question about $G_\delta$ set I'm on summer break but I want to keep my math skills sharp so I'm self-studying a bit from Munkres. 
This question is from pg 194, chapter 4 about the Countability and Separation Axioms. 
I've already seen this:
why is one point set in a first countable $T_1$ space a $G_\delta$?, but was wondering if my solution is also valid. 
$a.$ A $G_\delta$ set in a space $X$ is a set $A$ that equals a countable intersection of open sets of $X$. Show that in a first-countable $T_1$ space, every one-point set is a $G_\delta$ set. 
$b.$ There is a familiar space in which every one-point set is a $G_\delta$ set, which nevertheless does not satisfy the first countability axiom. What is it? 
Here's my attempted solution for $a.$ and I'm wondering if it's correct. 
$a.$ Let $\{x\}$ be a one-point set in $X$. Since $X$ is first countable there is a countable basis of $x$. Denote this by $\mathscr{B}$. 
WTS: $\cap_{n\in N}B_n$ = $\{x\}$. 
$\subset )$ $x\in\cap_{n\in N}B_n$ since each $B_n$ is a neighborhood of $x$. 
$\supset )$ Let $y \in \cap_{n\in N}B_n$. Then we have two cases.
Case 1: $y=x$ then we are done.
Case 2: $y\neq x$. Then $y$ is a limit point of $\{x\}$ since $y \in \cap_{n\in N}B_n$ and each $B_n$ contains $x$. Therefore $y \in \overline{\{x\}}$ but $\{x\}$ is closed since $X$ is $T_1$, so  $\overline{\{x\}} =\{x\}$, thus $y=x$. 
Therefore every one-point set in a first-countable $T_1$ space is a $G_\delta$ set. 
b. I'm quite unsure about $b.$ since it has to be a non-metric space. Any hints here would be great. 
 A: Your idea for a) is essentially OK: we take a countable local base $B_n$, $n \in N$ at $x$, and prove that $\cap_n B_n = \{x\}$. Indeed, clearly $x$ is in this intersection, as all $B_n$ must contain $x$. And if $y \neq x$, by $T_1$-ness, there is an open subset $O$ that contains $x$ but not $y$, so there is some $B_k$ with $x \in B_k \subset O$, so $B_k$ also does not contain $y$, and certainly $\cap B_n$ then does not. This shows that no other point than $x$ can be in $\cap B_n$.
For a familiar example I would pick $R^\omega$ in the box topology. If $x = (x_0, x_1,\ldots)$ is a point in it, the open sets $U_n = (x_0 - \frac{1}{n}, x_0 + \frac{1}{n}) \times (x_1 - \frac{1}{n}, x_1 + \frac{1}{n}) \times \ldots$ have the property that $\cap_n U_n = \{x\}$. But you can show that it is not first countable at any point (assume we have a countable local base at $x$, assume we have box-open basic sets as a local base WLOG, and then find a basic open set around $x$ that cannot contain any of them...).
Stefan's example of $\mathbb{R}/\mathbb{Z}$ will also do fine (it's less extreme, as it is first countable except at one point).
