Strict distance decrease on a compact metric space ($d(fx,fy)<d(x,y)(x\ne y)$)

A self-map $f$ of a nonempty compact metric space satisfying the displayed strict distance inequality has exactly one fixed point, although a uniform contraction constant need not exist. The inequality makes $f$ Lipschitz continuous with constant at most one, so $g(x)=d(fx,x)$ is continuous and attains a minimum at $p$. If $fp\ne p$, then $g(fp)<g(p)$, a contradiction. Two different fixed points contradict the same distance inequality.

For a strictly distance-decreasing map of a finite-dimensional Euclidean space, one bounded forward orbit suffices for existence of a fixed point. Its closure is compact by the Heine-Borel theorem and invariant by continuity, so strict distance decrease on a compact metric space applies. Strict distance decrease gives uniqueness throughout the whole space, not merely the orbit closure.