Due to results

we may assume that $X \times Y$ as well as $X$ and $Y$ are each nonempty.

Since $X$ and $Y$ are countable, then there exists surjections $f : \mathbb{N} \to X$ and $g : \mathbb{N} \to Y$. Define the map
\begin{equation}
h : \mathbb{N} \times \mathbb{N} \to X \times Y, \quad
h(n, m) : = (f(n), g(m))
\end{equation}
The goal is to show that $h$ is a surjection. To this end, fix $(x, y) \in X \times Y$. Since $f$ and $g$ are surjections, then there exists $n, m \in \mathbb{N}$ such that $f(x) = n$ and $f(y) = m$. Result

R3346: Characterisation of equality of ordered pairs implies that $(n, m) = (f(x), f(y)) = h(x, y)$ and hence $h$ is a surjection.

Result

R511: Natural number plane is countable shows that there exists a surjection $s : \mathbb{N} \to \mathbb{N} \times \mathbb{N}$. We can now form the composite map
\begin{equation}
h \circ s : \mathbb{N} \to X \times Y, \quad
(h \circ s)(n) = (f(n), g(m))
\end{equation}
whence result

R315: Composition of surjections is surjection guarantees that $h \circ s$ is a surjection. This finishes the proof. $\square$