Topology and axiom of choice It it an easy exercise to show that if $X$ is first-countable then for every point $x$ and every subset $A$ we have $x \in \text{cl}A$ iff there exists a sequence $(x_n)_n$ that converges to $x$.
Well, this uses the axiom of choice to create the sequence (I think). What would happen if we don't have that? (I know that in topology it is much better to have AC but I want to figure out what happens).

It is consistent with the ZF axioms that there is a dense set of reals $D\subset\mathbb{R}$ having no countable subset. Such a set is infinite, but Dedekind finite. It follows that any point in $\mathbb{R}-D$ is in the closure of $D$, but not a limit of any sequence from $D$, since any such sequence would give rise to a countable subset of $D$.
Meanwhile, your argument does not require full AC, but only countable AC, since you are making countably many choices of points closer and closer to $x$. 
