Is a set $U$ consisting of the single point $p$ open or closed? I'm guessing here that $U$ would have to be closed, especially since for example the proof of the theorem that the union of two closed sets is closed is also valid if one of the sets is $U$. Still, I'd like to make sure my approach is correct:

Since p is the single point in U, that means $\exists r&gt;0, B_r(p)\subset U$ can never be true, so $p$ is not an interior point, which means it has to be a boundary point, which means since $p$ is the only point in $U$, $U$ contains all its boundary points which means $U$ is closed.
I assume you're in a general metric space $(X,d)$. Consider {x} as a subset of $X$, and consider a point $y \neq x $ in $X$ . What can you say about d(x,y) (is it larger than 0)?  If so, can you construct a ball $B(y,r)$, for all $y \neq x $ that does not intersect the set {$x $}? (Remember that the complement of a closed set is open.)
let (X,d) be a metric space. we show X-{p} is open. let x belong to X-{p} and define r=d(x,p) then B(x, r/2) is a subset of X-{p}. in actually we show that each point in X-{p} is interior and  complement of {p} is open.this means {p} is close.