How to prove $\{y\in\mathbb R^n: |x-y|=r \text{ for some }x\in X\}$ is closed for closed $X$ and fixed positive $r$? Let $X\subset\mathbb R^n$ be a closed set and $r$ a fixed positive real number. Let $Y=\{y\in\mathbb R^n: |x-y|=r \text{ for some }x\in X\}$. Show that $Y$ is closed.
I tried to approach this problem with showing $Y^c$ is open, but I am stucked on how to use that $X$ is closed?

Suppose that $y_\omega$ is a limit point of $Y$; that is, for all integers $i \geq 1$, there exists some $x_i \in X$ and $y_i \in Y$ such that $y_i \in B_{1/i}(y_\omega)$ (open ball of radius $1/i$ and centered at $y_\omega$) and $|x_i - y_i| = r$.
Note that the sequence $y_1, y_2, \dotsc$ is bounded by $B_1(y_\omega)$, so the sequence $x_1, x_2, \dotsc$ is bounded by $B_{1 + r}(y_\omega)$. By the Bolzano-Weierstrass Theorem, there is a subsequence $x_{k_1}, x_{k_2}, \dotsc$ that converges to some $x_\omega$. Since $X$ is closed, $x_\omega \in X$. Finally, note that $|x_\omega - y_\omega| = r$ so $y_\omega \in Y$.
Since $Y$ contains every limit point $y_\omega$, it must be closed.
