I know that my direction for this proof is correct, but it feels like my reasoning for few things are off because I am still trying to grasp the Schroder-Bernstein so I would appreciate the guidance! Please note that I am novice on a good day..
Let A be an infinite set and B a countable set. Prove $|A|=|A\cup B|$
Suppose $B\subset A$, meaning union of A and B are disjoint, this results in $|A\backslash B|=|A|=|A\cup B|$ according to our claim. This also means that $A\backslash B$ is also infinite, since A is infinite. Let C be a denumerable subset of $A\backslash B$, and since B is countable and C is denumerable their union must also be denumerable. This means that $|B\cup C|=|C|$ such that there is a bijection $f:B\cup C\rightarrow C$. Now suppose we split A into two pieces, i.e. $A\backslash (B\cup C)$ which is the same as $(A\backslash B)\backslash C$. We already have a bijection $f:B\cup C\rightarrow C$, and we can form another bijection $g:(A\backslash B)\backslash C\rightarrow (A\backslash B)\backslash C$ meaning to itself. Let h be the function that is $f\circ g$, which by definition of composite functions, another bijection such that $h:A\rightarrow A\backslash B.$
I think you might be overcomplicating things, but your idea of using the Cantor–Schröder–Bernstein theorem is a good one. To apply the Cantor–Schröder–Bernstein theorem, you need to find an injection $A \to A \cup B$ and an injection $A \cup B \to A$. Well:
The inclusion function $i : A \to A \cup B$, given by $i(a)=a$ for all $a \in A$, is an injection.
Fix an enumeration of $B$, say $B = \{ b_0, b_1, b_2, \dots \}$. Since $A$ is infinite, it has a countably infinite subset $A' = \{ a_0, a_1, a_2, \dots \}$. Define $f : A \cup B \to A$ by letting $$f(a)= \begin{cases} a & \text{if } a \in A \text{ but not } A' \text{ or } B \\ a_{2n} & \text{if } a=a_n \in A' \text{ but not } B \\ a_{2n+1} & \text{if } a=b_n \in B \end{cases}$$ It is easy to prove that this function $f$ is injective.