= Solution
Let $K=\overline{O(x)}$. It is nonempty, closed and bounded in finite-dimensional <Euclidean space>, hence compact by the <Heine-Borel theorem>. The strict distance inequality again makes $f$ <Lipschitz continuous>. If $y\in K$, choose a sequence $y_j\in O(x)$ converging to $y$. Then $fy_j\in O(x)$, and continuity gives $fy_j\to fy$, so $fy\in K$. This proves \b[$f(K)\subseteq K$].
The restriction to $K$ satisfies <strict distance decrease on a compact metric space>, so part (b) produces a <fixed point> there. If a second <fixed point> existed anywhere in $\mathbb R^n$, the same strict distance inequality applied to the two points would contradict equality of their distance. \b[Therefore the fixed point is unique in all of $\mathbb R^n$.] This is the <bounded-orbit fixed point for a strictly distance-decreasing Euclidean map>.
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