Set containing open ball and contained in closed ball has the same boundary I'm having some difficult time trying to figure out how to prove that if $U$ is an open unit ball and $\overline{U}$ is a closed unit ball in $\mathbb{R}^n$ and $U\subseteq A\subseteq \overline{U}$ then the boundary of $A$ is the same as the boundary of $U$ or $\overline{U}$. Would definitely appreciate a hint.
My attempt:
$\mbox{bd}(\overline{U}) = \overline{U}\setminus U \supseteq \overline{U}\setminus A \supseteq \overline{U}\setminus U \supseteq \overline{U}\setminus \overline{U} = \emptyset$, so that $\overline{U}\setminus A = \mbox{bd}(\overline{U})=\mbox{bd}(U)$, which implies that bd$(A)= \mbox{bd}(U)$.
Unfortunately, my argument does not appear to be convincing enough.

We have that $cl(A)  \subseteq cl(B(0,1))$ from the fact that the closure of the set $A$ is the smallest closed set that contains $A$.
Also $B(0,1) \subseteq A \Rightarrow cl(B(0,1)) \subseteq cl(A)$
Thus $cl(B(0,1))=cl(A)$
Also  $int(B(0,1))=B(0,1)=int(A)$
Thus $bd(B(0,1))=cl(B(0,1))$ \ $B(0,1)=cl(A)$ \ $int(A)=bd(A)$
