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?

When you think you're defining g 'explicitly' , well, in fact you are not. You can do such 'explicit' definitions'  e.g. when you give an explicit formula on real numbers because you're manipulating only one or a few symbols (usually just 'x').
Here you're manipulating an arbitrary number of symbols $f_i$. Just because you think of them as indexed by i does not mean you really can refer to them all... until you have an axiom which says that you can do so. That's AC. It allows you to use an arbitrary number of symbols in that specific situation. In fact, it's more of a workaround than really using arbitrary number of symbols.
Remember that g is an object of set theory, which must be constructed using axioms. There's no such axiom as 'explicitly define a function'.
