The Cantor (1/3) set $C$ is constructed as follows:
$ C_0=[0, 1] $, $ C_1=[0, 2/3] \cup [2/3, 1] $, $ C_2=[0, 1/9] \cup [2/9, 1/3] \cup [2/3, 7/9] \cup [8/9, 1] $, and so on.
Then $ C= \bigcap_{i=1}^{\infty} C_i $. Also note that $ C_{i+1} \subset C_i $.
In many sources it is stated that any open cover of $C$ is an open cover of sets $ C_n, C_{n+1}, C_{n+2}, \dots $ when $n$ is "sufficiently" large. I guess that it might be intuitive but how could I justify it rigorously?
Suppose, toward a contradiction, that $\mathcal U$ is an open cover of $C$ but not of any $C_n$. So we have, for each $n$, a point $x_n\in C_n$ that is not in any member of $\mathcal U$. The bounded sequence $(x_n)$ has a convergent subsequence, say with limit $y$.
Each $C_n$ contains not only $x_n$ but also all the later terms $x_{n+k}$ (because $C_{n+k}\subseteq C_n$), and therefore contains all but finitely many terms of our convergent subsequence, and therefore contains the limit $y$ (because $C_n$ is closed). So $y\in C$ (by definition of $C$), and therefore $y\in X$ for some $X\in\mathcal U$ (as $\mathcal U$ covers $C$).
But $\mathcal U$ is an open cover, so $X$ is open and therefore contains all points sufficiently close to $y$. Thus, infinitely many of the $x_n$ are in $X$. But this is absurd, as all the $x_n$ were chosen to not be in any member of $\mathcal U$$.