Starting with two bijections,
$\quad f:\mathbb{Z^+} \rightarrow A, \; g:\mathbb{Z^+} \rightarrow B$
we can use recursion to define a function $h:\mathbb{Z^+} \rightarrow A \cup B$ as follows:
Set $h(1) = f(1)$.
Assume $h$ has been defined on the segment $[1,n)$ with $n \ge 2$ and define
$$
h(n)=\begin{cases}
f\bigr(\,\text{min}(\,f^{-1}\text{<}A \setminus h[1,n)\,\text{>})\,\bigr)\quad\text{if }n \text{ is odd}\\
g\bigr(\,\text{min}(\,g^{-1}\text{<}B \setminus h[1,n)\,\text{>})\,\bigr)\quad\text{if }n \text{ is even}\\
\end{cases}
$$
So by recursive definition we've specified our function $h:\mathbb{Z^+} \rightarrow A \cup B$.
One can use induction to show that $h$ is injective.
One can use induction to show that the range of $h$ includes $A$.
One can use induction to show that the range of $h$ includes $B$.
So $h$ is a bijection.
The basic idea is exactly what your instructor said. You have one big theorem, from the sounds of things:
If $A$ is denumerable (countably infinite), then $A \cup \{x\}$ is denumerable.
In other words, if you have a countably infinite set, it's still countably infinite after "adding one more thing."
So, when you have denumerable $A$ and finite $B = \{b_1, b_2, \ldots, b_n\}$, the basic idea is to create a bunch of intermediate denumerable sets. For example,
$$A \cup \{b_1\}$$
must be denumerable, by that theorem. Then, since $A \cup \{b_1\}$ is denumerable, if we add another element from $B$,
$$A \cup \{b_1, b_2\} = \left( A \cup \{b_1\} \right) \cup \{b_2\}$$
the theorem again applies, so that $A \cup \{b_1, b_2\}$ must be denumerable.
We could continue on and write out $A \cup B = A \cup \{b_1, \ldots, b_n\}$ as a bunch of repeated unions, but that gets clunky fast. However, it's exactly the idea behind using induction; we'll just make the argument a little cleaner.
Specifically, we'll let $A_k$ be the set $A \cup \{b_1, \ldots, b_k\}$, where $1 \leq k \leq n$. We want to show that $A_n = A \cup \{b_1, \ldots, b_n\} = A \cup B$ is denumerable.
Base case
$A_1 = A \cup \{b_1\}$ is denumerable. This follows directly from the theorem you were given.
Inductive Hypothesis:
$A_k = A \cup \{b_1, \ldots, b_k\}$ is denumerable. This is what we get to assume, to show that...
Conclusion:
\begin{align*} A_{k + 1} &= A \cup \{b_1, \ldots, b_k, b_{k+1}\} \\
&= \left(A \cup \{b_1, \ldots b_k\}\right) \cup \{b_{k+1}\} \\
&= A_k \cup \{b_{k+1}\}\end{align*}
is denumerable. I haven't explicitly said why it must be, but hopefully you can use your theorem to justify that $A_{k+1}$ is indeed denumerable, given that $A_k$ is.
If you can show that our inductive hypothesis implies the conclusion we want, then PMI will imply that $A \cup B$ is denumerable, for denumerable $A$ and finite $B$, as desired.
Best Answer
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.