Cardinals of set operations without AC Given info: $|A|=\mathfrak{c}$ , $|B|=\aleph_0$ in  ZF (no axiom of choice).
Prove: $|A\cup B|=\mathfrak{c}$
If $B \subset A\implies|A \backslash B|=\mathfrak{c}$?
I have found several places proving that for $|\mathbb{R} \backslash \mathbb{Q}|,$ but none of the solutions appears to work for arbitrary sets. Maybe one should prove it for $|\mathbb{R} \backslash \mathbb{Q}|$ and then show that it will work for arbitrary $A$ and $B$ as well?
Here Showing that $\mathbb{R}$ and $\mathbb{R}\backslash\mathbb{Q}$ are equinumerous using Cantor-Bernstein David has a very nice idea (constructing bijection), but it requires that infinite set has countably infinite subset, which again!? needs some choice axiom. 
In some  sources I even saw statements that these can't be proved in ZF.
From what my teacher said I percieved that solution has something to do with Cantor–Bernstein theorem and that knowing how to prove $\mathfrak{c}+\mathfrak{c}=\mathfrak{c}$ would help as well.
Thanks!

Prove it first for disjoint $A$ and $B$, relaxing the condition on $B$ to $|B|\le \aleph_0$. You then recover the full statement by considering $A\cup B = A\cup(B\setminus A)$.
You can restrict your attention even further to, say $A=(0,1)$ and $B$ being a subset of the integers. Once you have proved it for that case, the definition of "same cardinality" guarantees that it will be true for every other choice of disjoint $A$ and $B$ of the appropriate cardinalities.
It is true without any choice axiom that a set of size continuum has a countably infinite subset. By definition, because it has size continuum, there's a bijection from $\mathbb R$, and the image of $\mathbb N$ under that bijection is a countably infinite subset.
