So I'm trying to prove that the Cantor set $C$ is totally disconnected - that is, for any two points $x, y \in C$ there is a $z \notin C$ where either $x < z < y$ or $y < z < x$. I'd like to check if my argument proves this proposition.
First I assume without loss of generality that $x < y$.
Next, we fix $k \in \mathbb{N} \cup \{0\}$. If $|x - y| > 1/3^k$ we can easily argue that $x$ and $y$ are in two different subintervals of $C_k$ and choosing $z \notin C$ is straightforward.
If $|x - y| \leq 1/3^k$, then the two points are in the same subinterval of $C_k$. But since $x \neq y$, then we can choose a $1 \geq \delta > 0$ such that $|x - y| = \delta$. Now all we have to do is choose $n > 0$ such that $|x - y| = \delta > 1/3^{k + n}$. With some rearrangement we have $n > - k - \log_3(\delta)$ which means if we choose $n = -k + \lceil - \log_3(|x - y|)\rceil + 1$, then the set $C_{k + n} = C_{\lceil - \log_3(|x - y|)\rceil + 1}$ will result in $x$ and $y$ not being in the same subinterval. So there would exist a $z \notin C$ between $x$ and $y$.
How does my proof look?
A minimal choice of $k$ exists because the sequence $3^{-k}$ converges to $0$, i.e., there is an integer $K$ such that for all $k\geq K$, we have $3^{-k}<|x-y|$. By choosing any such $k\geq K$, we can argue that $x$ and $y$ are in different sub-intervals without cases.