Sincere request, don't forget to address my doubt at the end of the proof
I have assumed my sets to be disjoint at first but I have also addressed the general scenario as the proof progresses.
Set $A$ is said to be countable if there exists a bijection from $A$
to $\mathbb{N}$. Every countable set is infinite
To show that : Union of two countable sets is countable
Suppose $A$ and $B$ are countable.
Assume at first that $A\cap B=\phi$
$A $ countable $\Rightarrow \exists f:A\to \mathbb{N} $ a bijection.
$B $ countable $\Rightarrow \exists g:B\to \mathbb{N} $ a bijection.
define. $h:A\cup B \to N$ as
$x\mapsto 2f(x) \; $ if $x\in A$
$x\mapsto 2g(x)+1$ if $x\in B$
Because $A\cup B$ is infinite, it is sufficient to show that $h$ is injective in order to show that $A\cup B$ is countable.
if $x=y$, where $x,y\in A\cup B$, since $A$ and $B$ are disjoint, so, either both $x$ and $y$ belong to $A$ or both belong to $B$, and because $f$ and $g$ are well defined, so is $h$
Now let $h(x)=h(y)$ where $x,y \in A\cup B$
again, $x$ and $y$ can both belong to $A$ or can both belong to $B$. Hence injectivity of $h$ on $A\cup B$ follows directly from the injectivity of $f$ and $g$ on $A$ and $B$ respectively
Hence, $A\cup B$ is countable.
Now, let $A$ and $B$ be arbitrary countable sets,
then by above method, $A\cup B = [A\setminus (A\cap B)]\cup[A\cap B]\cup [B\setminus (A\cap B)]$ is countable.
Doubt : Is it safe to assume $A\cap B = \phi$ in the beginning of the proof? I am doubtful here because $A$ and $B$ are countable. Please address this problem first
Best Answer
Your proof is just about fine. It's perfectly acceptable to start with the case $A \cap B = \emptyset$ as long as you later address the possibility that $A \cap B \neq \emptyset$, which you do.
The one quibble is that you only proved that the union of two disjoint countable sets is countable, yet the end of your proof requires you to use this result for the union of three disjoint countable sets. (I assume you've already separately proved that any subset of a countable set is countable, which is how you know that each of your three components really is countable.) It's a tiny step, but for completeness you probably should show why the result for a union of three (or any finite) number of disjoint sets follows immediately from the result for a union of two disjoint sets.
Edited to add: Tuvasbien is correct to note that the final step of your proof can be simplified. $A \cup B = A \cup (B \setminus A)$, and $A \cap (B \setminus A) = \emptyset$.