The answer is given and up until a certain point it is clear. I would very much appreciate if anyone could help out understanding the solution after that point.
$$S \subset \mathbb R^2 \ \ \ \ |S|=\lambda_0 \ \ \ \ |\mathbb R^2|=c \ \ \ \ S \subset \mathbb R^2 \implies \exists x_0\in \mathbb R^2 \setminus S$$
Let $p$ be a line that contains point $x_0$. Let $H_1,H_2$ be a half-planes that define $p$.From this point is where I have troubles, all the way to the end:
$$p(x_0,\overline{x}), \overline{x}\in S, \theta \in [0,2\pi] \ \ \overline{x}\leftrightarrow\theta \\ |[0,2\pi]|=c \\ \ \overline{x}=\theta\overline{x} \ \ \ \ \ \{\theta\overline{x}|\overline{x}\in S\}= \lambda_0 \\ \{\theta\overline{x}|x\in S\}\subset[0,2\pi] \implies \exists \overline{\theta}\in [0,2\pi]\setminus \{\theta \overline{x}| \overline{x} \in S\}$$
Here's an alternative proof. Denote by $\mathcal{L}$ the set of horizontal lines in the plane. Define the function $f: S \to \mathcal{L}$ by associating to each point $x \in S$ the horizontal line $\ell \in \mathcal{L}$ passing through that point. You know that $S$ is countable and $\mathcal{L}$ uncountable.
Thus $f$ is not surjective (or else $card(S) \geq card(\mathcal{L})$). This means that there exists a horizontal line which does not contain any point in $S$.