Continuous surjections onto $\mathbb{R}$
I have two questions about continuous functions:
Suppose $X \subseteq \mathbb{R}$ and $X$ has same cardinality as $\mathbb{R}$. Can we find a continuous function from $X$ onto $\mathbb{R}$?
Suppose $X \cup Y = \mathbb{R}$. Can we find a continuous function from one of $X, Y$ onto $\mathbb{R}$?
The answer to (2) is affirmative. Let $f:R \to R^2$ be a continuous surjection. Let $\pi_x, \pi_y: R^2 \to R$ be projections onto $x, y$ axes. Now notice that if $X \cup Y = R$, then either $(\pi_x \circ f)[X] = R$ or $(\pi_y \circ f)[Y] = R$. Hence one of $X, Y$ can be continuously mapped onto $R$.
To make question (1) more interesting replace $\mathbb{R}$ by $[0, 1]$ so that the question becomes: Can we map every continuum size set of reals onto $[0, 1]$? Under CH, the answer is no by a simple diagonalization (or just note that a Lusin/Sierpinski set is a counterexample). One the other hand, Arnold Miller has proved (see here) that in the iterated Sacks model this is true. He also mentions an observation of J. Isbell: Whenever a perfect set of reals is partitioned into countably many sets, one of these sets can be continuously mapped onto $[0, 1]$.