Showing a quotient space is not first countable given instructions 
Let $X = (0,1] \times (0,1) \subset \mathbb{R}^2$ and $A= \{1\} \times (0,1)$. Show that the quotient space $X/A$ is not first countable. (Hint: let $\{B_n \mid n \in \Bbb N\}$ be a neighborhood base for $A \in X/A$ and $V_n = p^{-1}(B_n)$, where $p: X \to X/A$ is the quotient map. Construct a sequence $(x_n)$ such that $x_n \in V_n \setminus A$ and $x_n \to (1,1)$. Then $(X/A) \setminus\{p(x_n) \mid n \in \Bbb N\}$ is a neighborhood for $A$ in $X/A$).

What is the trouble that comes with the point $A \in X/A$? For every other point $[x] \in X/A$ if $U$ is an open neighborhood of $[x]$, then $p^{-1}(U)=\{x \in X \mid [x] \in U \} = \{x \in X \mid \{x\} \in U \}$. Is it true that for any $[x] \ne A \in X/A$ I can find a neighborhood base?
Trying to follow the instruction if $\{B_n \mid n \in \Bbb N\}$ is a neighborhood base for $A \in X/A$, then defining $(x_n)=(1+1/n,1)$ do I have that $x_n \in  V_n \setminus A$? The issue is understanding here what $V_n = p^{-1}(B_n)$ is. How can I be sure that $x_n \in V_n \setminus A$?
I also know that if $\{B_n \mid n \in \Bbb N\}$ is a neighborhood basis for $A \in X/A$, then $\{p^{-1}(B_n) \mid n \in \Bbb N\}$ is a collection of open subsets of $X$ containing $A$.

Only $A$ causes trouble, because $(X/A)\setminus\{A\}=(0,1)^2$ is first (and even second) countable.
By contradiction, let $B_n,V_n$ be as in the hint (not necessarily open, and not necessarily satisfying $(1\pm1/n,n)\in V_n$). $V_n$ is a neighborhood of the subset $A$ hence contains  some $x_n\notin A$, $1/n$-close to the point $(1,1-1/n)\in A.$
By construction, $x_n\to(1,1)\notin X,$ so that $X\setminus\{x_n\mid n\in\Bbb N\}$ is an open subset of $X,$ hence $(X/A)\setminus\{p(x_n)\mid n\in\Bbb N\}$ is an open neighborhood of $A$. But it contains no $p(x_n)$ hence no $B_n,$ a contradiction.
