I made the following proof for an excercise regarding the union of two countable infinite sets, namely, that their union is countable. It would be of great help if anyone could give me some feedback regarding the argument (ranging from the quality of the argument to some improvements). Btw, thanks for your attention.
The proof is as follows:
Since $A_{1}$ and $A_{2}$ are countable, it follows that there are biijective functions $f:\mathbb{N} \to A_{1}$ and $g:\mathbb{N} \to A_{2}$. Let $B_{2} = A_{2}\backslash A_{1}$ and so $A_{1}\cap B_{2} = \emptyset$. Also, consider the following sets $S_{1} = \{n\in \mathbb{N}: f(n)\in A_{1}\}$ and $S_{2} = \{n\in \mathbb{N}: g(n) \in B_{2}\}$ since $B_{2} \subseteq A_{2}$. If $B_{2} = \emptyset$, then $A_{1}\cup A_{2} = A_{1}$ and so $f:\mathbb{N} \to A_{1}\cup A_{2}$ is a biijective function and $A_{1}\cup A_{2}$ is countable. Hence, we may assume that $B_{2}\neq \emptyset$. By the Well Ordering Principle, $s_{2} = \min (S_{2})$ exists. Note that $S_{1} = \mathbb{N}$ and $1\leq s_{2}$.
Therefore, there are two possible cases. If $B_{2}$ is finite with $k$ elements, then define the function $h:\mathbb{N} \to A_{1}\cup B_{2}$ by
$\begin{align*} h(n) \begin{cases} g(s_{1}+(n-1)), \text{ if }n\leq k\\ f(n-k), \text{ if }n>k. \end{cases} \end{align*}$
Therefore, $A_{1}\cup B_{2} = A_{1}\cup A_{2}$ is countable.
If $B_{2}$ is infinite, then define some function $h:\mathbb{N} \to A_{1}\cup B_{2}$ by
$\begin{equation*} h(n) = \begin{cases} f(n/2), \text{ if }n \text{ is even}.\\ g(s_{1} + (n-1)/2), \text{ if }n \text{ is odd}. \end{cases} \end{equation*}$
Thus, $A_{1}\cup B_{2} = A_{1}\cup A_{2}$ is countable.
