Consider the following lemma (sometimes called Sequence Lemma)
Let $X$ be a topological space, $A\subseteq X$ any subset and $x\in X$. If there is a sequence of points in $A$ converging to $x$, then $x\in\bar A$; the converse holds if $X$ is first-countable.
A proof of the converse exhibits an enumeration of a neighborhood basis of $x$, which exits due to $X$ being first-countable, and constructs a sequence of points $(x_n)_n$ such that $x_n\in A\cap (U_1\cap\dots\cap U_n)$. These finite intersections are non-empty since $x\in\bar A$ and thus consequently $A\cap U\ne\emptyset$ for all neighborhoods $U\subseteq X$ of $x$ (in fact, as $x$ is a limit point of $A$ there is some point $y\in A$ different from $x$ in every $U$; but this does not seem relevant here). The proof is easily completed by the definition of a neighborhood basis and the construction of $(x_n)_n$ (see here for example).
However, we have to somehow (non-trivially?) choose the elements of the sequence $(x_n)_n$. I am aware of the Axiom of Countable Choice which should suffice here instead of, say, the full Axiom of Choice. Defining $S_i=A\cap(U_1\cap\dots\cap U_i)$ for $i\in\Bbb N$ and letting $S=\bigcup_{i\in\Bbb N} S_i$ should do the trick (unless I am fundamentally misunderstanding something). I am not sure if we actually need choice since the sets $S_i$ form a chain and this might allow a neat way of chosing "trivially".
Is the Axiom of Countable Choice needed here, does a weaker choice principle suffice, or is there maybe no need for any kind of choice at all?
Thanks in advance!
