I think we use a result of Howroyd? (following Moran? who did
Let's do the real line or Euclidean case of complete metric space?). . Let $X$ be a complete ? metric space with Hausdorff dimension $\alpha$. \alpha < \infty$. Then of course $X$ is separable, as well.
We use a result of Howroyd  (following Marstrand  who did the real line). Let $0 < \beta < \alpha$. Then $H^\beta(X) = \infty$, the $\beta$-dimensional Hausdorff measure. By Howroyd's theorem ($H^\beta$ is semifinite), there is a Borel ? subset $A \subset X$ with $0 < H^\beta(A) < \infty$. Then since a finite Borel measure is regular, there is a Cantor set $B \subseteq A$ with $0 < H^\beta(B) < \infty$, so of course $B$ has Hausdorff dimension $\beta$.
More details to be added
J. M. Marstrand, where "The dimension of Cartesian product sets." Proc. Cambridge, Philos. Soc. 50 (1954) 198--202
J. Howroyd, "On dimension and the ? are now..existence of sets of finite positive Hausdorff measure." Proc. London Math. Soc. 70 (1995) 581--604