Show that a subset of $(\mathbb R^n,\|\cdot\|)$ is closed 
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?
 A: First things first:

Let C be a closed subspace of the normed linear space $\mathbb{R}^n$.

I think you mean subset (as in the title): every subspace of $\mathbb{R}^n$ is closed due to finite dimension of $\mathbb{R}^n$.
Now, back to the issue at hand, given $C$ closed subset of $\mathbb{R}^n$.
Let $\bar{y}$ be an accumulation point of $D$. We want to show that $\bar{y}\in D$.
We have $y_n \in D$ such that $y_n\to \bar{y}$ and $x_n\in C$ such that $\forall n$ $\lVert x_n-y_n\rVert =r$.
From triangular inequality follows that $\lVert a\rVert - \lVert b\rVert\le \lVert a-b\rVert$. Using this inequality we get
$$
r-\lVert \bar{y}-y_n\rVert=\lVert x_n-y_n\rVert-\lVert \bar{y}-y_n\rVert\le\lVert x_n-\bar{y}\rVert
$$
and of course
$$
\lVert x_n-\bar{y}\rVert\le\lVert x_n-y_n\rVert+\lVert y_n-\bar{y}\rVert=r+\lVert y_n-\bar{y}\rVert\,.
$$
Choosing an arbitrary $\varepsilon>0$, there is a value $n_0$ such that $\lVert y_n-\bar{y}\rVert<\varepsilon$ $\forall n>n_0$. It follows that
$$
r-\varepsilon<\lVert x_n-\bar{y}\rVert<r+\varepsilon\qquad\forall n>n_0
$$
in other words
$$
\lVert x_n-\bar{y}\rVert\to r \qquad(1)
$$
Now, we can apply Bolzano-Weierstrass to $\{x_n\}$ since it's a bounded sequence. In fact
$$
\lVert x_n\rVert\le\lVert x_n-\bar{y}\rVert+\lVert \bar{y}\rVert
$$
where the first addend is convergent and the second one is fixed.
Thus we extract a convergent subsequence $\displaystyle\{x_{n_k}\}$ with limit value $\bar{x}\in C$, since $C$ is closed.
From $(1)$ follows that $\lVert \bar{x}-\bar{y}\rVert=r$.
Hence $\bar{y}\in D$ i.e. $D$ is closed.
