Let $X$ be a separable complete metric space. Let $E$ be a dense subset of $X$ and fix $p\in X$.
I want to 'choose' an element for each $\overline{B(p,1/n)}\cap E$ in ZF. ($n\in \mathbb{Z}^+$)
Is it possible?
As you can see in the link; Constructing a choice function in a complete & separable metric space
It's possible to choose representatives for $\overline{B(p,1/n)}$ for each $n\in \mathbb{Z}^+$.
I assume that $E$ is a countable dense subset, in which case it is clear that you can do that.
Write $E=\{e_n\mid n\in\omega\}$ and let $e\in\overline{B(p,\frac1n)}\cap E$ be $e_n$ such that $n=\min\{k\mid e_k\in B(p,\frac1n)\}$. By the density of $E$ you know that $E\cap B(p,\frac1n)$ is non-empty.
If $E$ is not well-ordered then this need not be possible. Suppose that $D$ is a Dedekind-finite dense set of real numbers.
We want to find $p_n\in D$ such that $|p-p_n|\leq\frac1n$. Because $p\notin D$ for every $x\in D$ you have some $n$ such that $|p-x|>\frac1n$, so you cannot choose the same $p_n$ more than a finite number of times.
Therefore such sequence would give us a sequence of elements from $D$, which is a contradiction to Dedekind-finiteness.