discrete metric, both open and closed. I've checked several answers though, still don't understand last bit.

Taking radius r = 1/2 then every subset is singleton and it is open.

But then how do you deduce it is also closed?

Well, a subset is closed if its complement is open... but then 'every' subset is open hence its complement is empty then closed as empty is both open and closed....??

Thanks.

ps. I haven't learnt about topology space, only metric space.
Note that if every subset is open, then every subset is closed: Given $A \subset X$, then the complement $A^c = X \setminus A$ is a subset, therefore open, and $A^c$ open is equivalent to $A$ is closed.

If you want to be concrete, you can view the complement of a single point as the union of the balls of radius $1/2$ centered on $y$, as $y$ ranges across every point other than $x$.  This union is evidently the complement of $\{x\}$.
let A is subset of discrete metric space.then to show A is closed then A contains all of its limit point.if r=1 clousre of A =A let if c is a limit point which is outside of A let r1 choose such that r1=(min (c,ai_)then open sphere centered with radius r1 not intersect A then A is closed