The problem is to prove the following equivalence:
$$2^{\aleph_0}=\aleph_1 \iff \exists\ A,B\in\mathbb{R}^2:\\ \textrm{a)}A\cup B=\mathbb{R}^2 \\\textrm{b)}\ \forall\ \textrm{lines } l\ \textrm{parallel to x-axis, } |A\cap l|\leq\aleph_0 \\\textrm{c)}\ \forall\ \textrm{lines } l\ \textrm{parallel to y-axis, } |B\cap l|\leq\aleph_0$$
I think I know how to prove "$\impliedby$":
Let's assume the RHS and that $2^{\aleph_0}\geq\aleph_2$. Now let's take a sequence $\langle x_{\alpha}:\ \alpha<\omega_1\rangle$. Since $\bigcup_\limits{\alpha<\omega_1}|B\cap x_{\alpha}|\leq \aleph_1\cdot\aleph_0=\aleph_1$ there is such a $y$-coordinate that neither of $(x_{\alpha},y)$ is in $B$. But hence we obtain $\aleph_1$ point which all lie on the same line parallel to $x$-axis. Since $A\cup B=\mathbb{R}^2$ they all must be contained in $A$ $-$ a contradiction as $A$ has at most countably many of those points.
But what about the other implication? Do I contruct these sets by induction? Over all lines or something else? I haven't come up with anything yet, perhaps you could help me.