Background: this question was asked in the course "Introduction to topology and metric spaces"
Prove that there exists a continuous function $f\colon[0,1]\to\mathbb R$ such that that for every $0 \leq r < s \leq 1$, $f$ is not Lipschitz on $[s,t]$.
Since the question was not asked in calculus or any kind of analytical course, I'm pretty sure that they don't expect a constructive proof.
Since all continuous functions $f\colon[0,1]\to\mathbb R$ are bound, then it follows that the metric space of these functions (with the supremum metric function) is a complete metric space. So far my main direction was using Cantor's intersection theorem to find the desired function in the intersection of a series of closed sets. The main problem is that the set of all such functions which are not Lipschitz on a specific $[s,t]$ is not a closed set. Another problem would be ordering these sets so that they are well-ordered and contained in one another. Both these problems make me think that even if Cantor's theorem is the right tool, I'm going about it the wrong way.