Show that the Cantor set is nowhere dense

I want to prove that the cantor set is nowhere dense. First, a subset $A$ of $\mathbb{R}$ is said to be nowhere dense provided that for every open set $\mathcal{O}$ has an open subset that is disjoint from $A$.

Proof:

Since the Cantor set is closed, then the closure of the Cantor set is the Cantor set. Then, to prove that the Cantor set is nowhere dense, it is enough to show that the interior is empty. Observe that for a subset $A$ of a topological space $X$, the interior of $A$ is defined as the union of all open sets contained in $A$. Let $A = C$, the Cantor set and $X = \mathbb{R}$, notice that $C = \bigcap_{k\in\mathbb{N}}C_k$, where each $C_k$ is closed. Moreover, we have $$\forall k\in \mathbb{N} \quad C_k = \bigsqcup_{k\in\mathbb{N}}F_k$$
where each $F_k$ is the disjoint union of $2^k$ closed intervals, each of length $1/3^k$. We thus conclude that $C$ dont contain any open sets and thus $$\text{int}(C) = \emptyset.$$

Is this proof correct?

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.