Proving that if a set A is denumerable and a set B that is finite and a subset of A, then $A\setminus B$ is denumerable Background:
This comes from the book: INTRODUCTION TO
MATHEMATICAL
PROOFS
Charles E. Roberts, Jr.
Indiana State University
Terre Haute, USA
A Transition to Advanced
Mathematics
Second Edition
A set $A$ is denumerable if and only if $A\sim \mathbb{N}$. 
$A\sim B$ if and only if there is a one-to-one correspondence (bijection) from $A$ to $B$.
Theorem 7.15 - If $A$ is a denumerable set and $B$ is a finite set, the $A\cup B$ is a denumerable set.
Question:

Prove that if $A$ is a denumerable set and $B$ is a finite subset of $A$, then $A\setminus B$ is denumerable. 

Attempted proof - Note that 
$$\begin{align*}
A\setminus B &= A\cap B^c\\
&= (A\cup B)\cap B^c
\end{align*}$$
Borrowing from the comments below. Since $A\cup B$ is denumerable by theorem 7.15, and any subset of a denumerable set is denumerable this implies that $A\setminus B$ is denumerable.
We know from theorem 7.15 that $A \cup B$ is denumerable. I am just not sure how to show that the complement of the finite set $B$ with $A\cup B$ is also denumerable. 
 A: First, we observe that $A \setminus B$ is not finite, otherwise $A = A \setminus B \cup B$ would be finite. Now, $A \setminus B$ (being infinite) is either denumerable or non - denumerable. Since $A \setminus B \subseteq A$ and $A$ is denumerable, $A \setminus B$ cannot be non - denumerable (A subset cannot contain "more" elements than its superset). Hence, the only possibility is that $A \setminus B$ is denumerable.
A: Try this:
We have that there exists a bijection $f:A \to \mathbb{N}$
Then $g = f^{-1}:\mathbb{N} \to A$ is a bijection too
And $B = \{b_{1},...,b_{k}\} \subset A$
So, we have that for every $n \in \mathbb{N}$,  $g(n) = a_{n} \in A$
And $g(n_{i}) = b_{i}$ for some $n_{i}$ with $ 1 \leq i \leq k$
And suppose that $n_{1}<n_{2}< \dots <n_{k}$
We are going to create a bijection $h:\mathbb{N} \to$ $A$ \ $B$ 
Let $d \in \mathbb{N}\cup\{0\}$, $d = n_{k} - k$
We have a subset with $d$ elements $\{1,2,...,n_{k}\}$\ $\{n_{1},...n_{k}\} = \{c_{1},...,c_{d}\}$
Let $h(n) = g(c_{n})$ if $1 \leq n \leq d$, and $h(n) = g(n + k)$ if $n>d$
Then $h$ define a bijection between $\mathbb{N}$ and $A$ \ $B$
