How to prove with the axiom of choice that : Given a family of non empty sets $(A_n)_{n\in\mathbb N}$, there exists a sequence $(x_n)_{n\in\mathbb N}$ such that for any $n\in\mathbb N$, $x_n \in A_n$.
I don't know how to proceed... fixing $n$ then apply the axiom of choice to $A_n$ ?
Thanks in advance for answers :)
On
You do not need to apply the axiom of choice in order to choose from one $A_n$, or even from finitely many of them.
The axiom of choice is needed when you want to choose from all of the $A_n$'s simultaneously. Of course, this cannot be done by repeatedly choosing elements, since a proof is finite, and there are infinitely many $A_n$'s.
However, the axiom of choice states (in one of its formulations) that if there is a family of sets, all of them are non-empty, then there is a function choosing an element from each set. This is literally what you're asking for. However, the axiom of choice has many many, many, forms in the literature. So the exact details would depend on your formulation of the axiom.
The definition here expresses the axiom of choice as
$$ \forall X \left[ \emptyset\not\in X \implies \exists f:X\to\bigcup X\quad \forall A\in X\ (f(A) \in A) \right]. $$
By taking $X:=\{A_n \mid n\in\mathbb{N}\}$, since each $A_n$ is nonempty, we are given a function $f:X\to\bigcup X$ such that $f(A) \in A$ for every $A \in X$. Now set $x_n:=f(A_n)$ for each $n\in\mathbb{N}$ to obtain the desired sequence.
In fact, this only requires countable choice because in our case $X$ is countable.