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).
On
Some papers: Disasters in metric topology without choice
Continuing horrors of topology without choice
and references therein.
On
Note that depending on the way you've defined the topology and how you want to use it, you may need choice to get the countable base at each point in your space. Thus even if you have dependent (countable) choice, there may be subtleties.
For example, suppose we work in ZF+DC+AD. Then $\omega_1$ with the usual topology is first-countable and we can even exhibit a countable local base at each point $\aleph\in\omega_1$, namely the collection of half-open intervals $\{(\beta,\alpha] : \beta<\alpha\}$ --- we can even order this in order-type $\omega$. However, we cannot uniformly order all the bases in order-type $\omega$. That is, there is no function $f:\omega_1\times\omega\to P(\omega_1)$ such that $\{f(\alpha,n) : n\in\omega\}$ is a local base at $\alpha$. (Recall that AD implies that there is no sequence $\{C_\lambda\subseteq\lambda\}_{\lambda\in\omega_1}$ such that $C_\lambda$ is a cofinal subset of $\lambda$ with order-type $\omega$.
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 $x$.