I ran into something which seems like it should be true (and must be true because it is used in the proof of a theorem).
I've extracted the detail that I cannot quite verify.
Let $(x_{n})$ be a sequence in a first countable topological space $X$. Let $y$ be an accumulation point of the set $I:=\{x_{n}:n\geq 1\}$ (that is, every neighborhood of $y$ contains $x_{n}$ for some $n\geq 1$ with the property that $y\neq x_{n}$).
Then $y$ is a limit point of the sequence $(x_{n})$. That is, every neighborhood $U$ of $y$ has the property that $U\cap I$ is infinite.
Note: I'm not sure if First countability is required here, but it was a hypothesis in the theorem (and was indeed used elsewhere in the proof so it probably isn't required) so I included it just in case.
Why I am stuck: If $X$ is a metric space, the argument is simple, assume for a contradiction that the set $U\cap I$ is finite, then take a ball $B:=B(y,\epsilon)$ with $\epsilon$ equal to half of $min\{d(y,x_{n}) : x\in U\cap I, x\neq y\}$. Then $U\cap B$ is a neighborhood of $y$ which contains no points of $I$ except possibly $y$ itself, which is a contradiction.
But since I'm not dealing with a metric space, this argument does not apply.
But I think that I can come up with a more general argument if $\{x\}$ is a closed set for every $x\in X$ (and more importantly for me, its complement is open).
Is this normally a hidden assumption in topological setings? Or is there another argument?