I came up with this idea to show that the Cantor set lacks isolated points (i.e. every point is a limit point) but I'm not sure if it works.
Let $C$ be the Cantor middle third set. My goal is to show that for any $x \in C$ and $\varepsilon > 0$ there is a $y \in C$ such that $|x - y| < \varepsilon$.
We can write
$$C = \left\{ \sum_{i = 1}^{\infty}{\frac{a_i}{3^i} : a_i \in \{0, 2\}} \right\}.$$
So, let $x \in C$ such that $x = \sum_{i = 1}^{\infty}{a_i/2^i}$. For $\varepsilon > 0$, we find $k$ such that $2/2^k < \varepsilon$ and we take $y \in C$ where $y = \sum_{i = 1}^{\infty}{b_i/2^i}$ such that
$$b_i = \begin{cases} a_i & i \neq k \\ 2 - a_i & i = k \end{cases}.$$
So, we know $y \in C$ and $|x - y| = 2/2^k < \varepsilon$.
Does this work?