Is there a Bernstein subset $X$ of $\mathbb{R}$ such that no continuous map $f : X → [0,1]$ is surjective?

Question

Is there a Bernstein subset $X$ of $\mathbb{R}$ such that no continuous map $f : X → [0,1]$ is surjective ?

Answer 1

No, if $X$ is a Bernstein set then there is a continuous surjection $X \rightarrow [0,1]$.

To see this, note that there is a continuous surjection $f: \mathbb R \rightarrow [0,1]$ such that the preimage of every point is an uncountable closed set. (For example, $f$ could be the composition of a space filling curve $\mathbb R \rightarrow [0,1]^2$ followed by a projection map $[0,1]^2 \rightarrow [0,1]$.) If $X \subseteq \mathbb R$ is a Bernstein set, then $X$ meets $f^{-1}(x)$ for all $x \in [0,1]$. Hence $f \restriction X$ is a surjection.