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Trivial "No" to all: Take the union $S$ of two disjoint balls $B_1$ and $B_2$ of diameters $d$ and $\sqrt{1-d^2}$ respectively. If $f$ maps $U$ to the ball $B$ of diameter $1$, then $f(B_1)$ has diameter $\le d$. If $d$ is small enough, then $B\setminus f(B_1)$ still contains two opposite points on the circumference, so the diameter of it is still $1$ and $f(B_2)$ has no chance to get anywhere close to covering it.