Let $C$ be a non-empty convex compact set of $\mathbb{R}^n$. Let $r>0$ and $U=C+B(0,r)$, where $B(0,r)$ is the open ball of radius $r$.
Prove that $Fr(U) \cap C = \emptyset$, where Fr(U) denotes the frontier of U.
I believe that one way of doing it is to show that the distance between $Fr(U)$ and $C$ is positive, in this case $r$.
I take a $y \in Fr(U), \; y= \bar{x} + u, \: u \in D(0,r)$ where $D(0, r)$ denotes the disc of radius $r$, then
$\forall x \in C$, $\| x - y\| \geq \inf_{x \in C} \| x-y\| = \| \bar{x} - (\bar{x} + u) \| = r$
and therefore $\operatorname{dist}(C,\operatorname{Fr}(U)) = r$
Is that correct? Thanks.
Your claim that $\inf_{x\in C}\lVert x-y\rVert=\lVert \overline x-(\overline x+u)\rVert$ seems unwarranted. Also, by "disk", do you mean "sphere" (say, $\{x\in\Bbb R^n\,:\, \lVert x\rVert=r\}$)? If so, it seems to me that proving that a point in $\operatorname{Fr}U$ is in the form $\overline x+u$ for some $u$ of norm $r$ and some $\overline x\in C$ is more important than the rest.
A different idea: If $V$ is open and $X$ is a subset, then $X+V$ is open, therefore we have $$C=C+\{0\}\subseteq U=\operatorname{int} U.$$ Thus $C\cap \operatorname{Fr} U\subseteq\operatorname{int} U\cap \operatorname{Fr}U=\emptyset$.