Often, we use a construction like this: Given a subset $ A $ of a metric space and its limit point $ a $, we know that for every $ \epsilon > 0 $ there is another point $ x $ different from $ a $ such that $ d(x,a) < \epsilon $. So for every natural $ n $ we let $ a_n $ be such a point within $ 1/n $ of $ a $, which gives us a sequence convergent to $ a $.
My question is, is the axiom of countable choice needed for this construction, or is it formally correct just in ZF?
This result is not provable in ZF. Cohen's original model for the failure of the axiom of choice satisfies the statement "There is a dense subset $D$ of the real line such that any sequence $(a_n)_{n\in\mathbb N}$ of elements of $D$ has only finitely many distinct elements."
Not only that the result is not provable in $\sf ZF$ itself. If we allow general metric spaces, then this is in fact equivalent to the axiom of countable choice.
Given a countable family non-empty sets, $X_n$, and let $A_n=\prod_{k\leq n}X_k$. If we can choose $a_n\in A_n$ then we can define a choice function from the $X_n$'s as well by defining $x_n=a_n(n)$ (recall that $a_n$ is a choice function). In fact, even if we can only choose from an infinite subset $I$ of the $A_n$'s we are done, by defining $k(n)=\min\{k\in I\mid n\leq k\}$ and taking $x_n=a_{k(n)}(n)$.
Consider $A$ to be the space whose underlying set is $\bigcup A_n\cup\{\infty\}$, where $\infty$ is a point which is not in any of the $A_n$'s. We define the metric to be as follows: $$d(x,y)=\begin{cases}0 & x=y\\\frac1n & x=\infty,y\in A_n\\\frac1n & x\in A_n,y=\infty \\ 1 & x\neq y,x\neq\infty,y\neq\infty\end{cases}$$
Clearly $\infty$ is an accumulation point of $\bigcup A_n$, therefore there exists a sequence $r_n$ such that $d(r_n,\infty)\to0$. But this concludes the proof, because a sequence converging to $\infty$ must pass through infinitely many of the $A_n$'s, therefore we can define $k(n)=\min\{k\mid r_k\in A_n\}$ and then we can take $a_n=r_{k(n)}$, where it is a well-defined choice, and this is a choice from infinitely many of the $A_n$'s, therefore it defines a choice from the family of the $X_n$'s from which we originally started.