So, start with the set [0,1] of the real line. Remove the middle third, and keep removing the middle thirds of the remaining line segments as usual when making the Cantor set.
Each time you remove a 3rd, add the middle point of the line segment you just removed to another set, S (initially empty). In the end you have the Cantor set C, and another set, S. Are the points in S countable?
Yes, $S$ is countable. At stage $n$ you removed $2^n$ open intervals, so you added $2^n$ points to $S$. $S$ is therefore the union of countably infinitely many finite sets and is therefore countable.
Another way to prove this is to notice that the open intervals that are removed to form the middle-thirds Cantor set are pairwise disjoint, and each contains exactly one point of $S$. Let $\mathscr{I}$ be the set of removed intervals. For each $s\in S$ let $I_s$ be the member of $\mathscr{I}$ containing $s$, and let $q_s\in I_s\cap\Bbb Q$. Clearly the map from $S\to\Bbb Q:s\mapsto q_s$ is injective (one-to-one), so $|S|\le|\Bbb Q|$, and $S$ is therefore countable.