The cantor set $C$ is defined as
$$ A_0 := [0, 1], \quad A_{n+1} := \frac{1}{3}(A_n \, \cup \, (2+A_n)), \quad C:= \bigcap_{n\in\mathbb{N}_0}A_n \text{.} $$
I need to show that $C$ is perfect (every point in $C$ is a limit point). I strongly believe that my proof idea is right but I am stuck at the point where I need to show that each $A_n$ consists of $2^n$ closed intervals and that the boundary points of these intervals are in $C$.
Let $x \in C$, and let $S$ be an open interval containing $x$. Let $I_n$ be the closed interval in $A_n$ (of which there are $2^n$) that contains $x$. Choose $n$ large enough so that $I_n \subset S$. Let $x_n$ be an endpoint of $I_n$ such that $x_n \neq x$.
Now, by construction, $x_n \in C$ (you should verify this). Hence, $x$ is a limit point: an arbitrary open interval $S$ containing $x$ contains $x_n \in C$.