Let $(X, d)$ be a metric space and $p\in X$, $\delta>0$ be fixed. Let
$$A=\{q \in X : p \in X, d(p,q)>\delta\}$$
How to show that $A$ is closed?
I tried to show that directly by taking $A$'s complement and show it is open but failed. I know when union boundary of some set becomes closed set, but I don't know how to prove it.
$d$ is a distance function, $X$ is an arbitrary metric space (not euclidean space)
I am studying analysis first nowadays, so I want you to do not "jump" throughout all proof. Thanks.
Assuming that $p\in X$ and $\delta > 0 $ are fixed (as stated in the comments), it seems clear that $A$ could be open (in fact this is the more "natural" conclusion, unless you can provide more details). To see this, let $p=0\in \mathbb{R}^2=X$, and let $d$ be the usual metric on $\mathbb{R}^2$, namely, $d(x,y)=|x-y|=\sqrt{x^2-y^2}$. The set $A^c$ in this case is written
$$A^c=\{q\in \mathbb{R}^2 : |q| \le \delta \}$$
If $\delta=1$, we call $A$ the closed unit disc (or closed unit ball); and the complement of a closed set is open, so $A$ must be open. If you have any further suspensions, one can easily check that $A^c$ is indeed closed (I'll leave this as an exercise, or consult any introductory book on point-set topology).