Observe that ${^{2}({^{\omega}2})}\thickapprox{^{\omega}({^{\omega}2})}\thickapprox{^{\omega}2}$. Consequently, each Cantor set contains a family of $\mathfrak{c}$ pairwise disjoint Cantor sets. If the Cantor set $2^{\omega}$ contains at least one homeomorphic subset to $\mathbb{Q}$, then $2^{\omega}$ has at least $\mathfrak{c}$ homeomorphic subsets to $\mathbb{Q}$.
Now, why is there at most $\mathfrak{c}$ homeomorphic subsets to $\mathbb{Q}$?
And how can I prove that $2^{\omega}$ has at least one homeomorphic subset to $\mathbb{Q}$?
The first question is just a consequence of cardinal arithmetic: the number of subsets of size $\kappa$ of a set of size $\lambda$ is bounded above by the set of functions from $\kappa$ to $\lambda$, which is just $\lambda^\kappa$ - now observe that $(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0}$.
In the other direction, note that $(1)$ every countable dense subset of $\mathbb{R}$ is homeomorphic to $\mathbb{Q}$ (indeed, any two countable dense linear orders without endpoints are isomorphic), and $(2)$ the Cantor set is quite similar to $\mathbb{R}$ - specifically, $\mathbb{R}$ is the image of the Cantor set under a continuous map which is injective except on a countable set. This suggests that to find something $\mathbb{Q}$-like in the Cantor set we should start by considering arbitrary countable dense subsets of the Cantor set; can you find one, and determine whether it is homeomorphic to $\mathbb{Q}$?