Let $A, B$ be some sets such that $A\setminus B$ is infinite while $B$ countable or finite. Prove that $A\setminus B\sim A$

Question

The problem is the following :
Let $A, B$ be some sets such that $A\setminus B$ is infinite while $B$ countable or finite. Prove that $A\setminus B\sim A.$

I got stucked in this problem by trying the following :
First, as $A\setminus B$ and $A\setminus B\subseteq A.$ By lemma, I can assume that there is an injection $f$ such $f: A\setminus B\to A.$ Thus, now I need to find an injection from $A\to A\setminus B,$ so I can use the Schröder–Bernstein theorem, that states :
If there's an injection $f$ such $f: A\to B,$ and there's another injection $g$ such $g: B\to A,$ then there is a bijection from $A$ to $B.$
As $A\setminus B$ is inifinite, there must exist a set $C$ countable such $C\subset A\setminus B.$
Now I may write that $A= (A\setminus B)\cup(C)\cup(A\cap B).$ Since $A\cap B\subseteq B,$ and $B$ is countable or finite, then $A\cap B$ is countable or finite as well, that means there exists an injection $A\cap B\to\mathbb{N}.$
But $C$ is countable, then there exists an injection $l$ such that $l: A\cap B\to C.$
That is, I found an injection from $A\cap B\to C,$ but how can I find now an injection $A\to A\setminus B$ in order to prove by the theorem ?

Answer 1

$A\setminus B=A-A\cap B$.
If $A$ is countable , so is $A\setminus B$ as $A\setminus B\subseteq A$ and therefore map every $i$th element of $A\setminus B$ to $i$th element of $A$ to get a bijection.

If $A$ is uncountable then so is $A\setminus B$ because if not then $A=(A\setminus B)\cup (A\cap B)$ is countable which is a contradiction.
Since $A\setminus B$ is uncountable, it contains a countable (infinite) subset $C$.
$A\setminus B=(A\setminus B-C)\cup C$ and $A=(A\setminus B-C)\cup ((A\cap B)\cup C)= (A\setminus B-C)\cup X$, where $X=(A\cap B)\cup C$ is countable.
Define $f:A\setminus B\to A$ by $f(x)=x$, if $x\in A\setminus B-C$ and $f(x_i)=y_i\in X$ for $x_i\in C$. $f$ is one -one.