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?

Let $f:A \to \mathbb{N}$ be a bijection from $A$ to $\mathbb{N}$
Let $g:B \to \mathbb{N}$ be a bijection from $B$ to $\mathbb{N}$
Define $h:A \times B \to \mathbb{N}$ as follows:
$h(x,y)=2^{f(x)}\cdot3^{g(y)}$
