Can there be an uncountable set $S\subseteq\mathbb R$ such that for each subset $D\subseteq S$, there is a Borel set $U$ with $D=S\cap U$?

I'm asking merely out of curiosity, but I'll mention that this would imply $2^{\aleph_1}=2^{\aleph_0}$. This is a hopefully more interesting adaption of a recent too easy question.

Lusin's second continuum hypothesisorLusin's hypothesis. $\endgroup$