Let $D\subset \mathbb{R}^n$ be closed and convex and let $D\subset B$ where $B$ is some open ball.
Let $x\in \partial D$ then $\exists y \in \partial B$ s.t. $\operatorname{dist}(x, \partial B)=|y-x|$. Is it true that also $|y-x|= \operatorname{dist}(y,D)$? (From picture it looks true)
No. In $R^2$ let $D$ be the convex region bounded by $\partial D=E\cup F\cup G$ where $E=\{(u,u+1): u\in [-1,0], F=\{(u,-u+1): u\in [0,1]\}, G=[-1,1]\times \{0\}.$
Let $B=\{(u,v)\in R^2:u^2+v^2=4\}.$ So $\partial B=\{(u,v)\in R^2:u^2+v^2=4\}.$
If $(1/2,1/2)\ne x=(a,1-a)\in G$ then $y$ is the intersection of $B$ with the half-line $L=\{(ua,u(1-u)):u\geq 1\}.$
But $L$ is not perpendicular to the line-segment $F.$ And when $x$ is sufficiently close to, but not equal to $(1/2,1/2),$ the foot $x^*$ of the perpendicular from $y$ to the line that includes $F$ will satisfy $x^*\in F,$ and $d(y,x^*)<d(y,x).$