Proving a subset is closed 
Let $X$ be a metric space and let $A \subset X$ be an arbitrary subset. We define 
  $$A_r=\{x \in X : B_r(x) \subseteq A\}.$$ Prove that $A_r$ is closed for every radius $r$.

Maybe this is easy but I am totally stuck. First I've tried to prove it directly, i.e., take a  convergent sequence in $A_r$ and prove that its limit is in $A_r$. Well, I didn't got anywhere. Then I've tried to prove that $A_r$'s complement is open. $A_r$'s complement is the set $$X \setminus A_r=\{x \in X : B_r(x) \not\subseteq A\}$$ Given $x \in X \setminus A_r$, I have  to take $\epsilon$ such that $B(x,\epsilon) \subseteq X \setminus A_r$. Can anyone give me a hint?
I add the proof:
To prove $A_r$ is closed is equivalent to prove its complement is open. So let $X \setminus A_r$={$y\in X\mid \exists u\notin A, y\in B_r(u)$}. Take $\epsilon=r-d(y,u)$ and consider $B(y,\epsilon)$. Let $z \in B(y,\epsilon)$, $d(z,u)\le d(z,y)+d(y,u)<r-d(y,u)+d(y,u)=r$. It follows that $z \in B_r(u)$, so $B_r(u)$ is open. $X \setminus A_r=\bigcup_{u\in X-U}B_r(u)$, as union of open sets gives an open set, $X \setminus A_r$ is open, which implies $A_r$ is closed. 

The set $A_r$ is the complement of the set $\{y\in X\mid \exists u\notin A, y\in B_r(u)  \}$. Can you show that this set is open?
