Do I use the axiom of choice when I construct a sequence that converge? Let $E$ closed and $x\in E$. I want to show there is a sequence that converges to $x$. I do as follow :
We have that $$E_n:=E\cap \left[x-\frac{1}{n},x+\frac{1}{n}\right]\neq \emptyset,$$
for all $n$. So, for all $n\in\mathbb N$, there is $x_n\in E_n$. Therefore $|x_n-x|<\frac{1}{n}\to 0$ when $n\to \infty $, and thus, $(x_n)$ is a sequence of $E$ that converge to $x$.
Did I used the Axiom of choice to construct $(x_n)$ ? If yes, how ? (because I didn't construct a choice function, did I ?)
 A: The statement is trivially true, since the constant sequence $x_n=x$ converges to $x$. It's also been pointed out in a comment that if we exclude $x$ (as you may have intended to imply), then the statement is false.
However, the corresponding statement for open sets is true: If $E\subseteq\mathbb R$ is open and $x\in E$, then there is a sequence $\{x_n\}_n$ with $x_n\ne x$ for all $n\in\mathbb N$ that converges to $x$. One possible proof proceeds in analogy to your proof, so we can ask the same question about this.
You do use a choice function, and yes, you are using the axiom of choice. You know that there is some $x_n\in E_n$, but you don't have a specific one, and you're not specifying a way to choose one. Thus, you're assuming the existence of a choice function to choose one.
You can prove the same thing without using the axiom of choice, though. Since $E$ is open, there is some $n_0\in\mathbb N$ such that $x+\frac1n\in E$ for all $n\ge n_0$. This sequence lies in $E$ and converges to $x$, and you don't need to assume the existence of a choice function to define it.
Update:
On your last question: No, you didn't construct a choice function, you assumed that one existed. It's the other way around: If you construct a choice function, you don't need to use the axiom of choice. You use it when you assume the existence of a choice function without constructing one.
