Two theorems:
$(1)$ Countable Union of Countable Sets is Countable
$(2)$ Cartesian Product of Countable Sets is Countable
Linked are the formal proofs on Proofwiki.
I do not understand why they had to use the Axiom of Countable Choice (ACC) the first proof but settled with "from the definition of countable" in the second. I think that $(1)$ should depend on ACC iff $(2)$ depends on it.
On
I was also a little confused when I first saw this example. I think the situation is clarified by introducing the following terminology: say that a counted set is a countable set $S$ together with a bijection $\mathbb{N} \to S$. It is true, in ZF, that
The Cartesian product of two counted sets is counted.
Finite choice, which is true in ZF, then implies that a Cartesian product of two countable sets is countable. This in turn implies, again in ZF, that
A countable union of counted sets is counted.
From here, to pass to the claim that a countable union of countable sets is countable, we need to turn countably many counted sets into countable sets, and this requires some form of countable choice. But the above theorem is good enough for most applications anyway. You will rarely know that a family of sets is countable without being able to explicitly find bijections from each of those sets to $\mathbb{N}$.
On
The axiom of choice allows us to make many choices at the same time.
It is true, without the axiom of choice, that if $(A_i,f_i)_{i\in\Bbb N}$ is a sequence of countable sets, and $f_i\colon A_i\to\Bbb N$ is an injection, then $\bigcup A_i$ is countable. This is because we are given the injections into $\Bbb N$.
However in the general case, one has to choose such injection (often a bijection). These injections are not at all definable that we can just write down some (perhaps very complex) formula witnessing their existence, and indeed without the axiom of choice it is possible that there is a family of countable (or even finite) sets whose union is not countable -- exactly because we cannot choose the injections in a uniform way.
On the other hand, when we take the product $A\times B$ we only have to make two choices, $f\colon A\to\Bbb N$ and $g\colon B\to\Bbb N$ - two injections - which is provably possible without the axiom of choice. And we already know how to find a bijection between $\Bbb N$ and $\Bbb{N\times N}$ in a very definable way, so that is enough.
The main issue here is that when we take products, or even unions, of infinitely many sets then we often need some uniformity in the way we order these sets (be it well-order, or otherwise). The axiom of choice allows to prove that such uniformity exists, because we can choose an ordering for each set; but without the axiom of choice it is sometimes the case that we may end up in a universe where such uniformity does not exist, i.e. without some form of the axiom of choice (or a weaker form) we cannot prove a particular statement.
The difference is that in the second proof we need to make only two choices: for each of $S$ and $T$ we must choose an injection into $\Bbb N$. Making a finite number of choices does not require any axiom of choice. In the first proof, however, we must make infinitely many choices, since we must choose an injection of each of the sets $S_n$ into $\Bbb N$, and that does require some choice principle.