I tried to prove that they are equivalent:
There is no set $A$ such that $\aleph_0<|A|<2^{\aleph_0}$ and
Every non-empty subset $A$ of $\mathbb{R}$ is a surjective image of $\omega$ or $\mathbb{R}$ is a surjective image of $A$. That is, $$\forall A\subseteq \mathbb{R} \exists f : (A\neq\varnothing \to f : \omega \twoheadrightarrow A \>\vee\> f: A\twoheadrightarrow \mathbb{R})$$
I don't think they are equivalent without the axiom of choice, but I don't know how to start to prove the unprovability. Thanks for any help!