 Solution

For a uniform contraction from a comparison function on a bounded set, the closed set $M$ is complete and has diameter at most $2R$. Suppose first $R>0$. The limit condition supplies $r_0>0$ and $q_0<1$ such that $\psi (r)/r\le q_0$ for $0<r\le r_0$. On the remaining compact interval $[r_0,2R]$, if nonempty, the continuous ratio $\psi (r)/r$ attains a maximum $q_1<1$, since it is strictly below one at every point. Thus there is one $q<1$ with $\psi (r)\le qr$ for every distance $0\le r\le 2R$ arising in $M$.

Consequently $d(Tx,Ty)\le qd(x,y)$ throughout $M$. Banach's theorem applies and gives a unique fixed point in $M$. No linear structure or compactness of the ball is used; the compact object in the argument is only the scalar distance interval. If the closed ball is the singleton at radius zero, its self-map fixes that point. The estimate on the ratio near zero is what prevents the compact-interval supremum argument from approaching one at its missing endpoint.