Let $C$ be a closed subspace of the normed linear space $(\mathbb R^n,\| \cdot \|)$.Let $r(>0)\in \mathbb R$
Define $D:=\{y:\exists x\in C$ such that $\|x-y\|=r\}$. Show that $D$ is closed.
EDITS:
Let $y_0\in D$,then $\exists x_0\in C$ such that $\|x_0-y_0\|=r\implies y_0\in B[x_0,r]\setminus B(x_0,r)$ where $B[x_0,r]=\{y_0:\|x_0-y_0\|\leqslant r\}$ and $B[x_0,r]=\{y_0:\|x_0-y_0\|< r\}$.
Thus $D=\cup_{x_0\in C} B[x_0,r]\setminus B(x_0,r).$
But the above characterization of $D$ fails to give any thing significant.What to do now?