Union of two countable set is also countable This question is asked many time when I search it. But i didn't find what am I really looking for.
My approach: 
Let $A_1,A_2$ be two countable sets. If they have some common elements then redefine it by $B_2=A_1\setminus A_2=\{x\in A_2:x\notin A_1 \}$. The point of this is that the union $A_1 \cup A_2=A_1 \cup B_2$ and the sets $A_1,B_2$ are disjoint. I guess three case be happen for $B_2:$


*

*If $B_2=\emptyset$, then $A_1\cup A_2=A_1 \cup B_2=A_1$ which we already know to be countable.

*If $B_2=\{b_1,b_2,\cdots,b_m \}$ has $m$ elements, then how can I define and make sure $h:A_1 \cup B_2\to \mathbb{N}?$ In order to satisfy the statement $A$ is countable iff there is a bijective $(1-1$ and onto$)$ mapping $f: A \to \mathbb{N}$, where $\mathbb{N}$ is the set of all natural numbers.

*If $B_2$ is inﬁnite again the same problem faced.


I have no idea how to approach and fix all these things. Besides, I wanted to know Is this approach is enough to proof the statement or not? Any hint or solution will be appreciated.
Thanks in advance.

Sets are countable if you can enumerate the elements ($=$ assign an index to each), without omission.
Let $C:=A\cup B$. From the enumerations
$$a_1,a_2,a_3,\cdots$$ and $$b_1,b_2,b_3,\cdots,$$
we form the enumeration
$$a_1,b_1,a_2,b_2,a_3,b_3,\cdots$$
also written
$$c_1,c_2,c_3,c_4,c_5,c_6,\cdots$$
In other terms, the elements of $A$ get the odd indexes and those of $B$ the even ones. You can check that this is an enumeration without omission.

If $A$ and $B$ have common elements, they will be cited twice. But this doesn't matter. (If you want, you can just skip the duplicates, but it is not even required.)
