The case $A=\emptyset $ is trivial, so we may dispose of it. Based on the formulation of what you are trying to prove, it seems the meaning of denumerable for you is that of a set that is either finite or has the same cardinality as $\mathbb N$. So, prove first that a set $S\ne \emptyset $ is denumerable if, and only if, there exists a surjection $f:\mathbb N \to S$.
Now, if $g:A\to B$ is surjective and $A$ is denumerable, then there exists a surjection $f:\mathbb N \to A$. The composition of surjective functions is surjective, thus $g\circ f:\mathbb N \to B$ is a surjection, so $B$ is denumerable.
You could do it all at once, but there’s nothing wrong with the idea of first proving that $X=A\cup B$ is countably infinite and then going on to prove that $X\cup C$ is as well. However, you’re off on the wrong foot in trying to prove that $X$ is countably infinite. Pairing up $A$ and $B$ isn’t going to help: what you need to do is pair up $A\cup B$ and $\Bbb Z^+$.
Writing $A=\{a_1,a_2,a_3,\dots\}=\{a_n:n\in\Bbb Z^+\}$ and $B=\{b_1,b_2,b_3,\dots\}=\{b_n:n\in\Bbb Z^+\}$ is a good start. Now let $X=A\cup B$. Informally you have $X=\{a_1,b_1,a_2,b_2,a_3,b_3\dots\}$, and it’s intuitive clear that this can be paired up with $\Bbb Z^+$:
$$\begin{array}{cc}
1&2&3&4&5&6&7&8&9&10&\dots\\
a_1&b_1&a_2&b_2&a_3&b_3&a_4&b_4&a_5&b_5&\dots
\end{array}$$
The problem is to express this clearly enough so that if someone asks what $55$ corresponds to, you can actually answer. If you examine the table above closely enough, it’s apparent that even numbers in the top line match up with $b$’s, and odd numbers with $a$’s, and a little more examination reveals the rule: $2n$ matches up with $b_n$, and $2n-1$ matches up with $a_n$. You can now express this as a bijection from $\Bbb Z^+$ to $X$:
$$f(n)=\begin{cases}
a_{\frac{n+1}2},&\text{if }n\text{ is odd}\\\\
b_{\frac{n}2},&\text{if }n\text{ is even}\;.
\end{cases}$$
Added: There is just one possible problem here: if the sets $A$ and $B$ are disjoint, all is well, but this $f$ will not be a bijection if some $a_i$ is equal to some $b_k$, i.e., if $A\cap B\ne\varnothing$. To get around this, you should replace $B$ by $B\,'=B\setminus A$. If $B\,'$ is finite, let $C'=B\,'\cup C$, and use the argument below to show that $A\cup C'$ is countably infinite. And if $B\,'$ is countably infinite, use it instead of $B$ in the argument above.
Now you need to prove that $X\cup C$ is countably infinite. Since we now know that $X$ is countably infinite, there’s no need to keep the notational baggage of $a_n$’s and $b_n$’s: let $X=\{x_n:n\in\Bbb Z^+\}$ (or, more informally, $\{x_1,x_2,x_3,\dots\}$). If $C=\varnothing$, then $X\cup C=X$ is certainly countably infinite, so we can assume that $C$ is non-empty, say $C=\{c_1,\dots,c_m\}$. Now the easiest way to list $X\cup C$ systematically is $\{c_1,c_2,\dots,c_m,x_1,x_2,x_3,\dots\}$. Lay out the correspondence with $\Bbb Z^+$ as we did above:
$$\begin{array}{cc}
1&2&3&\dots&m&m+1&m+2&m+3&\dots\\
c_1&c_2&c_3&\dots&c_m&x_1&x_2&x_3&\dots
\end{array}$$
It’s not too hard not to write down a bijection $g$ from $\Bbb Z^+$ to $X\cup C$:
$$g(n)=\begin{cases}
c_n,&\text{if }1\le n\le m\\\\
x_{n-m},&\text{if }n>m\;.
\end{cases}$$
Best Answer
Your proof for the first statement is on the right track: you are on your way to showing that any denumerable set $A$, finite or not, is a subset of an infinite set $S$, by constructing a set $A$ containing only but not necessarily all of the elements of $S$.
For the second statement - there's no "proof" for it because it is not true for all infinite sets.
E.g., consider $\mathbb R$: it is certainly an infinite set, but it is uncountable, and hence not denumerable, so it cannot be the subset of any denumerable set. If it were a subset of a denumerable set $S$, then as a subset of $S$, it would contain only, and at most all, elements of $S$ . Which means it would contain at most a countable number of elements, which contradicts the fact that $\mathbb R$ is uncountable.