Why is $\mathbb{R}/{\sim}$ not first countable at $[0]$, where $x \sim y \Leftrightarrow x = y\text{ or }x,y \in \mathbb{Z}$?

Let $\{U_n:n\in\Bbb N\}$ be a countable family of open nbhds of $[0]$; I’ll find an open nbhd of $[0]$ that does not contain any of them as a subset, thereby showing that they do not form a local base at $[0]$ and hence that $X/\!\!\sim$ is not first countable at $[0]$.

Let $q$ be the quotient map. For each $n\in\Bbb N$ and $k\in\Bbb Z$ there is an $\epsilon_{n,k}\in(0,1)$ such that

$$U_n\supseteq q\left[\bigcup_{k\in\Bbb Z}(k-\epsilon_{n,k},k+\epsilon_{n,k})\right]\;.$$

For $k\in\Bbb Z$ let $\delta_k=\frac12\epsilon_{k,k}$, and let

$$V=q\left[\bigcup_{k\in\Bbb Z}(k-\delta_k,k+\delta_k)\right]\;;$$

clearly $V$ is an open nbhd of $[0]$, and I claim that $U_n\nsubseteq V$ for each $n\in\Bbb N$. To see this, fix $n\in\Bbb N$; $\delta_n<\epsilon_{n,n}$, so we may choose a real number $x\in(n+\delta_n,n+\epsilon_{n,n})$. But then $q(x)\in U_n\setminus V$, and hence $U_n\nsubseteq V$.

By the way, a simpler version of the same idea is to start with $[0,1)\times\Bbb N$, where $\Bbb N$ has the discrete topology, and identify the set $\{0\}\times\Bbb N$ to a single point.