This is not intended to ask for the proof, but just for you to please check my proof. Thank you.
Definition of the Cantor set $C$:
$C = C_0 \cap C_1 \cap \cdots \cap C_i \cap \cdots$
where
\begin{align} C_0 &= [0,1] \\ C_1 &= [0,1/3] \cup [2/3, 1] \\ C_2 &= \left[0, \frac{1}{3^2}\right] \cup \left[\frac{2}{3^2}, \frac{3}{3^2}\right] \cup \left[\frac{6}{3^2}, \frac{7}{3^2}\right] \cup \left[\frac{8}{3^2}, \frac{9}{3^2}\right] \\ &\vdots \end{align}
Prove that $C$ has no interior points.
We aim to show that there doesn't exist a ball $B$, with radius $r >0$, that is entirely contained in $C$.
proof
Let $x \in C_n$ be an endpoint of one of the intervals of $C_n$, then $x \in C$. Note that by the construction of $C$, that $C_n$ has equal length intervals all of length $\frac{1}{3^n}$. So, for any given $r > 0$, choose the integer $n$ so that
$$n = \left\lceil\log_3\frac{1}{r}\right\rceil + 1 \implies \frac{1}{3^n}< r$$
Now, the ball $B_r(x)$ has a radius larger than the length of any interval in $C_n$ and it contains $x$. Since $x$ is an endpoint of an interval in $C_n$, that interval's other endpoint $y$ must satisfy
$$\left|y - x\right| = \frac{1}{3^n} < r$$
Therefore, by the arbitrariness of $r$, the Cantor set has no isolated points. But furthermore, $[x,y] \subset B_r(x)$, and we construct $C_{n+1}$ by deleting the middle third of $[x,y] \in C_n$ which means that there exists points in $B_r(x)$ that are not in $C$. Again, by the arbitrariness of $r$ we conclude $C$ has no interior.
Alternate proof inspired by copper.hat and Berci
proof
The measure of each $C_n$ is
$$mC_n = \text{num_intervals}\cdot \text{len_intervals} = 2^n\left(\frac{1}{3^n}\right) = \left(2/3\right)^n \to 0 \quad \text{as} \quad n \to \infty$$
Since each set $C_n$ is a subset of $C_{n-1} \cap C_{n-2} \cap \cdots \cap C_1$, then
$$m(C_1 \cap C_2 \cap \cdots \cap C_n) = mC_n$$
and therefore $mC = 0$.
Now if $C$ contains any open set of the form $(a,b)$ then $mC \geq m(a,b) = b-a$. Since $mC = 0$, $C$ must not contain an open set, which implies it can't contain an open ball, which implies $C$ contains no interior points.