Countable collection of countable sets and Axiom of choice Do we need Axiom of choice(or weaker version axiom of countable choice) to say countable Cartesian product of countable sets is nonempty? I think yes.
I read somewhere answer no giving argument: each countable set can be well ordered and after well ordering each countable set we choose least element in each to prove their Cartesian product is non empty. But I see gap in this argument because there are many ways a countable set can be well ordered. So which way we will well order sets?

If you have a product $\prod_{i\in I}x_i$ of sets $x_i$ indexed by some set $I$, then an element in that product is exactly a choice of an element $y_i \in x_i$ for each $i \in I$. The axiom of choice (and its weaker counterparts) will therefore tell you exactly that such an element always exists.
Note that there are products that can be proven to be non-empty even without the axiom of choice. For instance, if all the $x_i$ are ordinals or subsets of ordinals.
And yes, choosing a well-ordering for each $x_i$ is (at least) as difficult as choosing a single element, logically speaking. So unless there's a distinguished well-ordering you can choose (for instance, if the $x_i$ are ordinals, then $\in$ defines a well ordering for all of them, or if the union $\bigcup_{i \in I}x_i$ admits a well-ordering, or $I$ is finite), then you're back where you started, and in need of choice.
