A proof that a countable product of countable sets is non-empty that does not use the axiom of choice. Is the proof correct? Let $I$ be some non-empty set of indexes and for every $i \in I$ let $A_i$ be a set with cardinality $\aleph_0$.
Is the following proof that the cartesian product $\prod_{i \in I}A_i$ is non-empty valid without the axiom of choice?
By definition, since $|A_i| = \aleph_0$ for every $i \in I$, then there are bijections $f_i: \mathbb{N} \rightarrow A_i$ for every $i \in I$. Let’s define explicitly the function of choice $g: I \rightarrow \bigcup_{i \in I}A_i$ by: $g(i) = f_i(0)$. It’s clearly a function of choice, hence $\prod_{i \in I} A_i \ne \emptyset$. QED.
If it is valid, can it be generalized to any cartesian product of sets with equal cardinality? If it is not valid, why?

Your argument does invoke choice, albeit in a subtle way: when you choose a family of bijections $\{f_i: i\in I\}$. Just because, for each $i$, the set $F_i$ of bijections from $A_i$ to $\mathbb{N}$ is nonempty, doesn't mean that you can pick one for each $i$; this is exactly the axiom of choice applied to the family $\{F_i: i\in I\}$. In order to define $g(i)$, you need to refer to a specific $f_i$, so this use of choice is not easy to remove from your argument; and in fact it can be proved that the statement you are seeking to prove is not provable in ZF (= set theory without choice) alone. 
In fact, this holds in the most powerful way possible: even choice for families of two-element sets is not provable in ZF!
