Let $D$ be the set of points in $\Bbb R^2$ such $\lvert p \rvert\leq1$, and let $f: D\rightarrow D$ be a surjective function, satisfying this relation: $\lvert f(p)-f(q)\rvert \leq \lvert p-q\rvert$, for all $p$,$q$ in $D$.
Proof that $f$ is an isometry, i.e.: $\lvert f(p)-f(q)\rvert = \lvert p-q\rvert$.
HINT:
Take $P$, $Q$ mapping to $P'$, $Q'$ a diameter, so $PQ$ must be a diameter. For $R$ on the segment $PQ$ we have $PR+RQ = PQ$ so the image $R'$ lies on $P'Q'$ at the same distance from the ends. This is true for the preimage of any diameter. Now take $S'$ on the circle, image of $S$ on the circle. Since $PS\ge P'S'$, $QS\ge Q'S'$ we must have equality throughout. Hence the transformation is a rotation or a symmetry.
In fact, taking $g\colon = f^{-1}$: if $g \colon K\to K$ a transformation of a compact metric space $K$ with $d(g(x), g(y) ) \ge d(x,y)$ then $g$ is an isometry.