Show that the space $\Bbb R^2 / \ell$ is not first countable. 
Consider the quotient space $\Bbb R^2/\ell$, where $\ell$ is a line through the origin in $\Bbb R^2$. Show that this space is not first countable.

Without loss of generality we can assume that $\ell = \Bbb R \times \{0\}$ i.e. the $x$-axis.
Now aiming for a contradiction if $\Bbb R^2/\ell$ was first countable we could find a neighborhood base $\{B_n \mid n \in \Bbb N\}$ for $\ell \in \Bbb R^2/\ell$.
These basis sets are in one-to-one correspondence with the open sets $\{\pi^{-1}(B_n) \mid n \in \Bbb N\}$ containing $\ell \subset \Bbb R^2$ where $\pi:X \to X/A$ is the quotient map.
Thus for any $B_n \in \{B_n \mid n \in \Bbb N\}$ we get an open set $\pi^{-1}(B_n)$ containing $\ell \subset \Bbb R^2$.
If we pick one such set say $B_{n_0}$, then I'm told to consider a set $\{(n, x_n) \mid n \in \Bbb N\}$ such that $x_n$ is a positive real number inside $\pi^{-1}(B_{n_0})$ now presumeably $\{(n, x_n) \mid n \in \Bbb N\}$ is closed so $\Bbb R^2 \setminus \{(n, x_n) \mid n \in \Bbb N\}$ is open  and contains $\ell$ and so $\Bbb R^2 \setminus \{(n, x_n) \mid n \in \Bbb N\} \in \{\pi^{-1}(B_n) \mid n \in \Bbb N\}$.
The questions are. What is the motivation for the set $\{(n, x_n) \mid n \in \Bbb N\}$ and why is this a contradictive result?
 A: The overall argument is ok, it needs a bit tweaking though. Lets make it fully correct.
We start with the countable family $\{B_n\ |\ n\in\mathbb{N}\}$ of neighbourhoods of $[\ell]$ in $X/\ell$. For a fixed $n\in\mathbb{N}$ we consider a single point $(n,x_n)$ such that $x_n\neq 0$ and $(n,x_n)\in \pi^{-1}(B_n)$. Since $\pi^{-1}(B_n)$ is an open neighbourhood of
$\ell=\mathbb{R}\times\{0\}$ then of course such $x_n$ has to exist. All together we are doing a diagonal-like argument: for $n$-th $B_n$ set we pick a special $(n,x_n)$ point with $n$ on the first coordinate.
Now $B=\{(n,x_n)\ |\ n\in\mathbb{N}\}$ is indeed closed. Pretty much because $\lVert (n,x_n)-(m,x_m)\rVert\geq 1$ for any $n\neq m$, and thus a convergent sequence in $B$ has to be eventually constant. This in fact holds regardless of what $x_n$ are, the particular choice will be important later. Therefore $V=\mathbb{R}^2\backslash B$ is open. And it is a neighbourhood of $\ell$, since all $x_n\neq 0$.
Now collapse relationships have this very interesting property:

Lemma. Let $X$ be a space, $A\subseteq X$ a subset and consider the quotient map $\pi:X\to X/A$. If $C\subseteq X$, then
$$\pi^{-1}\big(\pi(C)\big)=\begin{cases}
C &\text{if }A\cap C=\emptyset \\
C\cup A&\text{otherwise}
\end{cases}$$
In particular, if $A\subseteq C$ then $C=\pi^{-1}(\pi(C))$.

which I leave as an exercise.
This implies that $V=\pi^{-1}(\pi(V))$ because $\ell\subseteq V$. And thus $\pi(V)$ is an open neighbourhood of $[\ell]$ in $X/\ell$. But none of $B_n$ is contained in $\pi(V)$, that's precisely why and how we choose $(n,x_n)$ points.
