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!

By assumption, there exist bijections $a\colon A\to \mathbb R $ and $b\colon B\to\mathbb N_0$. 
We can define the injective map $b'\colon B\setminus A\to\mathbb N_0$, $x\mapsto\lvert\{y\in B\setminus A\mid b(y)<b(x)\}\rvert$.
If $b'$ is a bijection, we can define $c\colon A\cup B\to\mathbb R$ by
$$c(x)=\begin{cases}a(x)&\text{if }x\in A\text{ and }a(x)\notin\mathbb N_0,\\
2a(x)&\text{if }x\in A\text{ and }a(x)\in\mathbb N_0,\\
2b'(x)+1&\text{if }x\in B\setminus A.\end{cases} $$
If on the other hand $b'$ is not a bijection, let $N=\min(\mathbb N\setminus b'(B\setminus A))$. Then clearly $b'(A\setminus B)= \{k\in\mathbb N_0\mid k<N\}$ and $b'$ is a bijection with this set so that we can define
$c\colon A\cup B\to\mathbb R$ by
$$c(x)=\begin{cases}a(x)&\text{if }x\in A\text{ and }a(x)\notin\mathbb N_0,\\
a(x)+N&\text{if }x\in A\text{ and }a(x)\in\mathbb N_0,\\
b'(x)&\text{if }x\in B\setminus A.\end{cases} $$
In both cases, $c$ is readily seen to be a bijection.
