Not necessarily. If is the discrete metric, and , the statement fails.Wont the closure of A be { q in X | d(p,q) <=r}
and closure of B be {q in X | d(p,q) >=r }
You might be able to show that the closure is contained in there, however.
Fixp inX, r > 0, define A to be the set of all q inX for which
d(p, q) < r, define B similarly, with > in place of <. Prove that A and
B are separated.
Wont the closure of A be { q in X | d(p,q) <=r}
and closure of B be {q in X | d(p,q) >=r }