Bijection from $A$ to $S\setminus A$, where $A$ is countably infinite Original question:

Suppose $S$ is an infinite set such that $|N|\leq|S|$
show there exists a countable infinite set $A$, where $A$ is a subset of $S$, and there is a bijection between $S\setminus A$ and $S$.

I know the same question is here but it doesn't conclude and it doesn't go down the path I end up taking.
My work,
I have proven there is a countably infinite subset of S. The proof I used is the same as proof 2 in this wiki proof guide.
This is where things get shaky. To begin, I think the way I constructed set $A$ makes my proof impossible.
I think there are two cases to consider. first $S$ is infinite, but not countable.
Let $S = T\bigcup A$, therefore $S\setminus A=T$
So, I'm am just showing a bijection form $T$ to $S$. I know $T$ is a subset $S$ and I know that $T$ is infinite and uncountable from $A\bigcup T$.
I don't know how to progress here.
Case 2, If $S$ is countable infinite then the way I constructed $A$ would mean that $S\setminus A = \{\emptyset\}$ as $A$ would be $S$.
P.S
errors were pointed out to me, I would like to amend some statements above without obfuscating what was originally said.
First, I meant to say proof 1 in the wiki article. I wanted to use the image of injective mapping $\phi:N\rightarrow S$ as my term for A
 A: Before beginning, I would just like to say that this is not my area of expertise but I can give a rough idea on what I believe the proof should look like. However, I am not familiar with what level of detail is required for these sorts of proofs so take what I say with a grain of salt.
Case $1$: $S$ is countable. If $S$ is countable then there is a bijection between $S$ and $\mathbb{N}$. Define $A$ to be the set of elements in $S$ that correspond to even numbers. That is
$$S=\{s_1,s_2,s_3,...\}$$
$$A=\{s_2,s_4,s_6,...\}$$
Now, there clearly exists a bijection between $S$ and $S\backslash A$ since both are infinite, countable sets (and we are done).
Case $2$: $S$ is uncountable. As you have noted, it is possible to get a countable subset $A\subset S$ (for proof, see here). Let $A$ be any countable subset of $S$ and define $T=S\backslash A$. Since $S$ is uncountable and $A$ is countable, we know $T$ is also uncountable. By the same logic, define $B$ to be any countable subset of $T$. Now, note that
$$S=A\cup B\cup (S\backslash (A\cup B))$$
$$S\backslash A = B\cup (S\backslash (A\cup B))$$
Note that there exists a bijection between $A\cup B \to B$ since both of these sets are infinite and countable. Denote this bijection with $\phi(a)$ for all $a\in A\cup B$. Now, let us describe the bijection $f:S\to S\backslash A$.  We have
$$f(s)=\begin{cases} 
      s & s\in S\backslash (A\cup B)\\
      \phi(s) & s\not\in S\backslash (A\cup B) 
   \end{cases}$$
Since these are both bijective maps, we conclude $f:S\to S\backslash A$ is a bijection.
