Let $X_1, X_2, X_3, \ldots$ be countable sets. Due to results

we may assume that all of $X_1, X_2, X_3, \ldots$ are nonempty. Consider the sets $Y_1, Y_2, Y_3, \ldots$ defined by
\begin{equation}
Y_N
: = \prod_{n = 1}^N X_n
\end{equation}
We proceed by induction on $N$. The base case of $N = 1$ is immediate since $X_1$ is assumed to be countable. As an induction hypothesis, assume then that $Y_N$ is countable for some $N \geq 1$. We must show that this implies that $Y_{N + 1}$ is countable. Since $Y_N$ and $X_{N + 1}$ are countable, result

R263: Binary cartesian product of countable sets is countable guarantees that the cartesian product
\begin{equation}
Y_N \times X_{N + 1}
= \left( \prod_{n = 1}^N X_n \right) \times X_{N + 1}
\end{equation}
is countable. Further, result

R1848: Bijection between parenthesis-sliced cartesian products shows that
\begin{equation}
|Y_N \times X_{N + 1}|
= |Y_{N + 1}|
\end{equation}
That is, $Y_{N + 1}$ is countable. The claim is now a consequence of

R800: Proof by principle of weak mathematical induction. $\square$