Q:

Is this function necessarily a contraction?

If $(X, d)$ is a compact metric space satisfying $d(f(x), f(y)) < d(x, y)$ for all $x, y \in X$ such that $x \ne y$, is $f$ necessarily a contraction? 
I know an analogue of the Banach Fixed Point Theorem applies in this case, and the proof is easier than that for $X$ merely complete. But is there a short proof that there is some $\alpha \lt 1$ such that $d(f(x), f(y)) < \alpha d(x, y)$ for all $x, y \in X$ such that $x \ne y$?
I imagine there is, because of the relationship between compactness and upper bounds, and $\alpha$ acts as an upper bound of sorts here. But I can't come up with the proof.

A:

No. Let $X=[0,\frac12]$ with the usual metric and $f(x)=x-x^2$.

