In a discrete topological space, all subsets are open by definition. Consider a fixed family B of subsets of the overall space X. Since all subsets the universal set X are open, their complements relative to the set are closed. But it's not hard to show that the complements of the complements of each set is the family B itself! So every set in B is also closed. So every subset of X is either open or closed.
Now the trick is to formulate this argument specifically for metric spaces in terms of open balls. Sketch of Proof: Formulate the family B as a nested sequence of open balls of equal constant distance c centered at an arbitrary point x Note c doesn't have to be 1,but make c=1 if it makes it easier for you. Demonstrate that in addition to being closed, the set of complements of open balls in B is itself a sequence of open balls in X. (This is a perfectly general argument since each metric space has a countable neighborhood basis of each point(local base),but you don't have to worry about this technicality for this argument.) Consider the possible values of c in the construction of the open ball sequence in comp(B). This will show each subset of comp(B) is either a singleton (which is open) or an open ball containing 2 points of diameter c. This will give you the proof.
Addendum:Clearly what I meant was that each metric space was first countable. Adam Smith,who loves to cherry pick my posts,pointed this out. I've corrected it and it was a fairly egregious error.