Cartesian Product of Countable Sets is Countable/Informal Proof

Proof
Let $S = \left\{{s_0, s_1, s_2, \ldots}\right\}$ and $T = \left\{{t_0, t_1, t_2, \ldots}\right\}$ be countable sets.

If both $S$ and $T$ are finite, the result follows immediately.

Suppose either of $S$ or $T$ (or both) is countably infinite.

We can write the elements of $S \times T$ in the form of an infinite table:


 * $\begin{array} {*{4}c}

{\left({s_0, t_0}\right)} & {\left({s_0, t_1}\right)} & {\left({s_0, t_2}\right)} & \cdots \\ {\left({s_1, t_0}\right)} & {\left({s_1, t_1}\right)} & {\left({s_1, t_2}\right)} & \cdots \\ {\left({s_2, t_0}\right)} & {\left({s_2, t_1}\right)} & {\left({s_2, t_2}\right)} & \cdots \\ \vdots & \vdots  & \vdots & \ddots \\ \end{array}$

This table clearly contains all the elements of $S \times T$.

Now we can count the elements of $S \times T$ by processing the table diagonally.

First we pick $\left({s_0, t_0}\right)$. Then we pick $\left({s_0, t_1}\right), \left({s_1, t_0}\right)$. Then we pick $\left({s_0, t_2}\right), \left({s_1, t_1}\right), \left({s_2, t_0}\right)$.

We can see that all the elements of $S \times T$ will (eventually) be listed, and there is a specific number (element of $\N$) to index each of its elements with.

Thus we have the required one-to-one correspondence between $S \times T$ and $\N$, and our assertion is proved.