Certainly there are irrationals in the Cantor set that can be described simply and explicitly, such as the number that has base $3$ expansion $0.200200020000200000200000020\dots.$ If the number were rational, its base $3$ representation(s) would be ultimately periodic. But it isn't, because of the increasing number of $0$'s between consecutive $2$'s.
Added: An interesting related question is whether there is a closed form irrational number in the Cantor set. The meaning of that question is not clear since we have not defined closed form. However, let $\alpha=\sum_0^\infty \frac{2}{3^{n(n+1)/2}}.$ Then $\alpha -2$ is an irrational number in the Cantor set, for basically the same reason as the example we gave in the main post. But $\alpha=\sqrt[8]{3}\; \vartheta(0, 1/\sqrt{3}),$ where $\vartheta$ is the Jacobi $\vartheta$-function. Unfortunately, $\vartheta$ is a pretty exotic function. If we define closed form more narrowly, I do not know what the answer would be.