Let $(K,d)$ be a compact metric space. Prove or disprove: There does not exist a continuous function $f:K \to K$ such that $d(f(x),f(y))>d(x,y)$ for all $x\neq y$.
My attempt: I tried using Brouwer's fixed point theorem but couldn't come up with a continuous function $g$ such that $d(g(x),g(y))<d(x,y)$ for all $x\neq y$.
Hint: think about $d$ as a function $K \times K \to \Bbb{R}$. It's continuous, and as $K \times K$ is compact, its image is compact and hence bounded. Again, using the fact that $K \times K$ is compact, $d$ must achieve its maximum value, i.e., for some $x_0, y_0 \in K$, $d(x_0, y_0) \ge d(x, y)$ for every $x, y \in K$. So $d(f(x_0), f(y_0)) \le d(x_0, y_0)$ for any function $f : K \to K$.