Show that the Cantor set is nowhere dense

The closure of the Cantor set is the same Cantor set, for it is closed. The interior of the Cantor set is empty, since it contains no interval. Thus, the Cantor set is nowhere dense: its closure has empty interior.

In addition to @Pedro's answer, and using the fact that the question is tagged in measure theory, there is a quick answer to why the Cantor set has empty interior.

By its construction, it is clear that $m(C)=0$, where $m$ is Lebesgue measure. If $C$ had non-empty interior, it would contain an interval $(a,b)$. But $(a,b) \subset C \implies m\big((a,b) \big) \leq m(C) \implies m\big((a,b) \big)=0,$ a contradiction. Note that the argument actually shows that every set with Lebesgue measure $0$ has empty interior.