The first claim is indeed true. The key to the proof is to use the following (equivalent) definition of denumerability ( = countably infinite )
Def: a set $X$ is denumerable if all of its elements can be enumerated in a sequence, i.e.all of its elements can be listed as a sequence $x(1), x(2), ...x(n)$, ...
So using this definition we can write a proof for this claim.
Suppose $A$ is a subset of $B$ and $B$ is denumerable. So we can enumerate the elements of $B$ as a sequence $x(1), x(2), ..., x(n),.$. The subsequence consisting of those elements that belong to $A$ is then an enumeration of $A$. We are done.
The second claim is not true. Take $A = \mathbb{N}$ and $B = \mathbb{R}$, then $A$ is a subset of $B$ and $A$ is denumerable while $B$ is not.
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
$A$ is countable iff there is a function $f : A \to \omega$ such that $f$ is one-to-one(not necessarily onto) if $A$ is finite.
Let $ Id_A : A \to A$ is one-to-one and one can make the image bigger without effecting one-to-one propriety $Id_A : A \to B$ and $ g: B \to \omega$ such that $g$ is one-to-one, then simply take $ f = g(Id_A) : A \to \omega$ and its not hard to prove that $f$ is one_to_one.
Assume $x,y \in A$ and $f(x) =f(y) $ and we need to prove that $ x =y$
$f(x) = g(Id_A(x)) = g(x) = g(y) = g(Id_A(y)) = f(y)$, which means that $ g(x) =g(y)$ which means that $x=y$ because we know that $g$ is one_to_one.