Sequence Lemma explanation

Question

Then every neighbourhood $U$ of $x$ contains a point of $A$.

So I don't see it happening unless $X$ is a metric space, but the proof is for any topological space.

Answer 1

That follows immediately from the definition of convergence of a sequence.

Let $X$ be any topological space, $\sigma=\langle x_k:k\in\Bbb N\rangle$ a sequence in $X$, and $x\in X$. Then by definition $\sigma$ converges to to $x$ if and only if for each nbhd $U$ of $x$ there is an $n_U\in\Bbb N$ such that $x_k\in U$ for all $k\ge n_U$.

In your case we’re assuming that $\langle x_n:n\in\Bbb N\rangle\to x$, so by definition every nbhd $U$ of $x$ not only contains at least one $x_n$, but actually contains all terms of the sequence from some point on. Since by hypothesis the sequence lies entirely in $A$, every nbhd of $x$ therefore must contain at least one point of $A$.