Countable Union of Countable Sets is Countable/Proof 2

Theorem
The union of a finite or countable number of countable sets is countable.

Proof
Consider the countable sets $$S_0, S_1, S_2, \ldots$$ where $$S = \bigcup_{i \in \mathbb{N}} {S_i}$$.

Assume that none of these sets have any elements in common.

Otherwise, we can consider the sets $$S_0^\prime = S_0, S_1^\prime = S_1 - S_0, S_2^\prime = S_2 - \left({S_0 \cup S_1}\right), \ldots$$

All of these are countable by the fact that they are subsets of countable sets, and they have the same union $$S = \bigcup_{i \in \mathbb{N}} {S_i^\prime}$$.

Now we write the elements of $$S_0', S_1', S_2', \ldots$$ in the form of an infinite table:

$$\begin{array} {*{4}c} {a_{00}} & {a_{01}} & {a_{02}} & \cdots \\ {a_{10}} & {a_{11}} & {a_{12}} & \cdots \\ {a_{21}} & {a_{21}} & {a_{22}} & \cdots \\ \vdots & \vdots  & \vdots & \ddots \\ \end{array} $$

where $$a_{ij}$$ is the $$j$$th element of set $$S_i$$.

This table clearly contains all the elements of $$S = \bigcup_{i \in \mathbb{N}} {S_i}$$.

Now we can count the elements of $$S$$ by processing the table diagonally. First we pick $$a_{00}$$. Then we pick $$a_{01}, a_{10}$$. Then we pick $$a_{02}, a_{11}, a_{20}$$.

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

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