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.

Say that $U=B(0,1)$. Since $\overline U$ is closed, we have that $\overline A\subset\overline U$. Since $U$ is an open ball, every point of $U$ is an interior point of $A$. Hence, $\partial A\subset \overline U\setminus U=\{x:\,|x|=1\}=\partial U$. Taking $x\in\partial U$, for every $r>0$, $B(x,r)$ intersects both $U\subset A$ and $\mathbb{R}^n\setminus \overline U\subset \mathbb{R}^n\setminus A$ and so $x\in \partial A$.
