I wonder if this is true:
If $X$ is a separable metric space of cardinality $|\mathbb R|$, then $X$ contains a Cantor set.
What if, additionally, $X$ is perfect (no isolated points) and zero-dimensional? (Assume this if it makes the argument easier).
Here's a partial answer, where we don't impose a dimension hypothesis:
Using the axiom of choice, we can construct a set of reals that - as a subspace of $\mathbb{R}$ with the usual topology - doesn't contain any Cantor sets. The key point is that there are only continuum-many closed sets of reals, and in particular only continuum-many Cantor sets in $\mathbb{R}$, and so we can do a standard diagonalization argument to produce such a set.
In fact, we can do even better: we can construct a Bernstein set, which is a set of reals which meets every perfect set but contains no perfect set. Note that the complement of a Bernstein set is also a Bernstein set. Furthermore, Bernstein sets have to be dense (so no isolated points).
Conversely, over ZF the determinacy of the perfect set game - and hence a fortiori the axiom of determinacy - implies that the answer to your question is yes (even without a dimensionality hypothesis).