I like the $|x-y| >3^{-n}$ proof for this, but also wonder whether my proof below is acceptable.
proof: Assume that the Cantor set $P$ is not totally disconnected. i.e. it contains a subset $S$ that is connected.
Since S is bounded, by the least upper bound property (and also g.l.b.p), $a = inf(S)$ and $b=sup(S)$ exist. Since S is connected, the interval $(a,b)\subset S$. So $S$ contains an interval and thus $P$ contains an interval. Contradiction.
