Prove that if $A$, $B$ are countable, then $A \times B$ is countable? Is $A\times B$ referring to the axis here? So an $X$ and $Y$ coordinate plane?
$A$ is countable, therefore a bijection occurs from $A \rightarrow \mathbb{N}$.
$B$ is countable, therefore a bijection occurs from $B \rightarrow \mathbb{N}$.
If these statements hold true then any elements belonging to the sets can be linked and ordered one-by-one from any natural number, thus any coordinate $(a,b)$ would be countable as the elements it represents ($x$ and $y $ coordinates) are countable. Together they preserve their nature of cardinality.
Am I wrong?

Another possibility is along the following lines.
Given enumerations
$$
a_1, a_2, a_3, a_4, \ldots
$$
and
$$
b_1, b_2, b_3, b_4, \ldots
$$
of the sets $A$ and $B$ respectively, you can enumerate the set $A\times B$ of pairs as follows
$$
(a_1,b_1),(a_2,b_1),(a_1,b_2),(a_3,b_1),(a_2,b_2),(a_1,b_3),(a_4,b_1),(a_3,b_2),(a_2,b_3),(a_1,b_4),...
$$
(the pattern should be clear).
