With a cursory examination, it looks like the OP's proof is in the ballpark. They asked for a proof that would be more efficient, so here we will highlight the main ideas found in the OP's proof, presented as a logical progression.
In what follows we do not assume that our sets are contained in $\mathbb R$.
Proposition 1: If $A_1$ and $A_2$ are two disjoint countably infinite sets, then the union $A_1 \cup A_2$ is countably infinite.
Proof:
Let $F_1: \mathbb N \to A_1$ and $F_2: \mathbb N \to A_2$ be bijective mappings, where $\mathbb N = \{0, 1, 2, \dots, n, \dots \}$. Define $F: \mathbb N \to A_1 \cup A_2$ as follows:
$$
F(m) = \left\{\begin{array}{lr}
F_1(\frac{m}{2}), & \text{for even } m \in \mathbb N\\
F_2(\frac{m+1}{2}-1), & \text{for odd } m \in \mathbb N
\end{array}\right\}
$$
It is easy to check that $F$ describes a bijective correspondence. $\quad \blacksquare$
The following sequence of 'add-on' propositions are not difficult to prove:
Proposition 2: Let $A_1$ and $A_2$ be any two sets. Then there exist a set $B$ satisfying the following:
$\tag 1 B \subset A_1 \text{ and } B \subset A_2$
$\tag 2 A_1 \cup A_2 \text{ is the disjoint union } (A_1 \backslash B) \cup B \cup (A_2 \backslash B)$
Proposition 3: If $A_1$ is a countably infinite set and $A_2$ is any finite set, then the union $A_1 \cup A_2$ is countably infinite..
Proposition 4: Any subset of a countably infinite set is either finite or countably infinite.
Proof (sketch)
Use the lemma found here.
Proposition 5: The union of a finite number of finite sets is finite.
We now prove the main proposition.
Proposition 6: The union of any two countably infinite sets $A_1$ and $A_2$ is countably infinite.
Proof:
By proposition 2, 4 & 5, $A_1 \cup A_2$ can be written as a disjoint union
$\tag 3 C_1 \cup B \cup C_2$
where each of the sets is either countably infinite or finite, and at least one of the sets is countably infinite.
By using the commutativity and associativity laws of the union operation and proposition 1 and/or proposition 3, we can simplify (3) in two steps so that a single countably infinite set remains that is equal to $A_1 \cup A_2$. $\quad \blacksquare$
The OP might want answer this question again at some point using the following:
Theorem: If $f$ is a surjective function from $\mathbb N$ onto a set $A$, then $A$ is either finite or countably infinite.
Best Answer
I would proceed like this:
We know that there is a bijection between $A_1$ and $\Bbb{N}$ and $A_2$ and $\Bbb{N}$. Let's call them $f_1: A_1 \longrightarrow \Bbb{N}$ and $f_2: A_2 \longrightarrow \Bbb{N}$.
We need to prove that there is $g: B \longrightarrow \Bbb{N}$ bijective ($B = A_1 \cup A_2$), or equivalently, a $g': \Bbb{N} \longrightarrow B$.
So define $g'$ as follows: $$g'(n) = \begin{cases} f_1^{-1}(n), & \text{if n is even} \\ f_2^{-1}(n), & \text{if n is odd} \end{cases}$$
Note that $f_1$ and $f_2$ are invertible because they are bijective. Now take $g = g'^{-1}$ and you have what you were looking for.
Note: a way to visualize it is to consider the following two sets: $A = \{(+, n) : n \in \Bbb{N}\}, B = \{(-, n) : n \in \Bbb{N}\}$ which union gives $\Bbb{Z} \setminus {0}$ and notice that $h$ defined as $$h(n) = \begin{cases} (+, n/2), & \text{if n is even} \\ (-, (n+1)/2), & \text{if n is odd} \end {cases}$$ is a bijection between $A \cup B$ and $\Bbb{Z} \setminus {0}$. And you know that $\Bbb{Z} \setminus {0}$ is countable.