Let $S$ be a relation over $\mathcal P(\mathbb R)$ such that $A,B\subseteq\mathbb R: \exists f:A\to B, \exists g: B\to A$ and $f,g$ are injections.
Find the cardinality of $\{X\in \mathcal P(\mathbb R)| XS\mathbb Z\}$.
We should note that $S$ is an equivalence relation and from CSB we know that $|A|=|B|$ so we can write the above set like so: $\{X\in \mathcal P(\mathbb R)| |X|=\aleph_0 \}$, from here I think this set is isomorphic to an infinite union of countable sets and we know that this is countable, but, it also represent $\mathcal P (\mathbb Q)$ because all the subsets of the reals that are countable are all the reals apart from the transcendental numbers. Probably the latter line of thinking is correct.
I should use the injections in order to prove the cardinality is $\mathfrak c$ but I can't seem to make sense of it.
Use the Cantor-Bernstein theorem. Then you first have to prove that there are at least $\mathfrak c$ countably infinite subsets of $\mathbb R$, and then that there at most $\mathfrak c$ of them.
The first one is easy -- the sets $\mathbb{Z}\cup\{x\}$ for $x\in(0,1)$ are $\mathfrak c$ different countably infinite subsets.
For the second one, you need a uniform, injective way to code each countable subset of the reals as a real. If you have the Axiom of Choice you can do that by showing that every sequence of reals can be encoded as a single real (which is a simple exercise in using the equalities $|\mathbb R|=|\mathcal P(\mathbb N)|$ and $|\mathbb N|=|\mathbb N^2|$).
I don't think the conclusion is necessarily true without Choice.