I previously had a question that was too long and should have probably been split into two questions. I was able to show that if $X$ was a complete seperable metric space with a subset homeomorphic to the Cantor set, then $\left[0,1\right]$ was the continuous image of $X$. I am having trouble showing this though:
If $X$ is a complete separable metric space and $\left\lvert X\right\rvert=2^{\aleph_{0}}$, then $X$ contains a subset homeomorphic to the Cantor set.
I know that since $\left\lvert X\right\rvert=2^{\aleph_{0}}$, then it has a set of condensation points, denoted as $\operatorname{cond}\left(X\right)$. I also know that $\operatorname{cond}\left(X\right)$ is perfect, so it is closed and thus is a complete perfect metric subspace of $X$ (also separable, since any subspace of a separable metric space is separable. Not sure if this is useful). My idea was to try and construct a homeomorphism $f$ from a closed subset of $\operatorname{cond}\left(X\right)$ to the Cantor set, but I've been having trouble coming up with one.