As the title suggests, I have to prove, if (X, d) is a metric space, that for any subset A of X, diam A = diam Cl(A), i.e. the diameter of A equals the diameter of its closure. So, if A is closed, it is trivial, since Cl(A) = A. Assume A is open. Now I'm a bit lost. If A is open in (X, d), then it is a union of open balls in X. By the way, I know that for every open ball K(x, r) in X, diam K = diam Cl(K). Also, I have shown that Cl(A) can be written as a union of closed sets in X (since the closure of a union equals the union of closures). Also, I know that, since A is a subset of Cl(A), diam A <= diam Cl(A) must hold. And also, I know that for subsets A, B of X, diam (A U B) <= diam A + diam B + d(A, B), which can be generalized for any finite number of subsets. The definitions of diameters didn't get me anywhere, too. For some reason, I can't make any use of these facts to prove it. Any suggestion?