I don't understand how the diameters of $ \bar A$ and $A$ are equal I'm given the assignment to prove that $\operatorname{diam}\bar A = \operatorname{diam}A$, where $\operatorname{diam}A=\sup\{\rho(a,b): a,b \in A\}.$
How can they be equal if $\bar A= \partial A \cup A$:
Say $x\in \partial A$, then does it not follow that $x > a, \; x> b$, thus $ x \notin A$ ?

We don't necessarily have any ordering among the points of the metric space (consider e.g. $\Bbb R^n$), so your statements $x>a$ and $x>b$ are meaningless. 
To prove the claim, take any $a, b\in\bar A$ and consider sequences $a_n, b_n\in A$ with $a_n\to a, \ b_n\to b$, and note that $\rho(a_n, b_n) \to\rho(a, b) $. 
