Existence of a sequence of elements picked arbitrarly 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 :)

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.
