This question is in the context of the following exercise: Let $X$ be a first countable topological space, let $A \subseteq X$, and let $x \in X$ with $x \in \overline{A}$. Then there exists a sequence $\{a_{n}\}_{n=1}^{\infty} \subseteq A$ with $a_{n} \rightarrow x$.
My work:
Let $\{U_{i}\}_{i \in \mathbb{N}}$ be a countable nbhd base for $x$. Consider the sequence $\{\cap_{i=1}^{n}U_{i}\}_{n=1}^{\infty}$. Each $\cap_{i=1}^{n}U_{i}$ contains an element of $A$. Call this element $a_{n}$. Then $a_{n} \rightarrow x$ because if $\mathcal{O}$ is an open nbhd of x, there's some $j$ such that $U_{j} \subseteq \mathcal{O}$. Then $a_{n} \in \cap_{i=1}^{n}U_{i} \subseteq \cap_{i=1}^{j}U_{i} \in \mathcal{O}$ for all $n \geq j$. My question is: is it o.k. to form a sequence in that way? Do we get the information of what the points $a_{n}$ are all at once? (in which case we have tangible objects with which we form our sequence) The axiom of choice wouldn't be needed, would it?
No, the axiom of choice needed. In fact the axiom of countable choice is equivalent to this exercise (it is equivalent in the case of metric spaces, which are first-countable).
The reason is that you chose $a_n$ from each set in the sequence. This means that you made infinitely many choices, but you haven't (and weren't) supplied with a uniform way of making them. Therefore the axiom of choice was invoked.
There is indeed a counterexample where the axiom of choice fails and there is a metric space which has a closure point which no sequence can reach.
Other than that, the proof seems fine.