I apologize that I cannot post the exact question and I know this is too vague, but I am asked to show that, around some point $x$ in some quotient space, it does not have a countable basis around that point. I was guessing that, then around $x$, it does have to have an uncountable basis. But I'm getting nowhere from here, and I cannot really think of any trick.
Maybe I need to use the contradiction, but I'm not sure what "having countably many basis" will imply. Could someone give me any input?
The commonest way I've seen is a kind of diagonalisation argument: given countably many $(B_n)_n$ open neighbourhoods of $x$ we can find an open neighbourhood $O$ of $x$ such that $B_n \nsubseteq O$ for all $n$.
You could also contradict a consequence of first countability: e.g. $x \in \overline{A}$ for some set $A$ while there is no sequence from $A$ converging to $x$.
It all depends on who the topology is defined. For quotients the first way is the most common, maybe because open sets are easy to recognise.