Is it consistent with ZFC that there exists an uncountable separable metric space $X$ such that every subset of $X$ is a Borel set?

If the continuum hypothesis holds, or more generally $2^{\aleph_{0}}<2^{\aleph_{1}}$ , then each uncountable separable metric space contains non-Borel sets since there are only $2^{\aleph_{0}}$ Borel sets. Therefore, in order for there to be an uncountable separable metric space $X$ where every subset of $X$ is Borel, we would need to at least have $2^{\aleph_{0}}=2^{\aleph_{1}}$. What other conditions are necessary in order for there to exist an uncountable separable metric space $X$ where every subset of $X$ is Borel?