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$ ?

You want to show  $$\operatorname{diam}\bar A = \operatorname{diam}A$$
Obviously $$\operatorname{diam}A \le\operatorname{diam}\bar A  $$
On the other hand if $$\operatorname{diam}A <\operatorname{diam}\bar A  $$ we have two points in the $\bar A$ whose distance is greater than $\operatorname{diam}A$ 
At least one of these points must be in  $\partial A $
That implies that the point in the boundary is included in an open ball which does not intersect A.
That contradicts the definition of  $\partial A $
