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!

Note that if $A$ has cardinality $\frak c$ then $A$ has a countably infinite subset. This does not require the axiom of choice. $|\mathcal P(\Bbb N)|=\frak c$ and $|X|<|\mathcal P(X)|$.
As for the suggestion to use the fact $\frak c+\frak c=\frak c$, this is again helpful because you can replace $B$ with its power set and you have: $$\mathfrak c=|A|\leq |A|+|B|\leq |A|+|\mathcal P(B)|=\frak c+c=c.$$
