Proof verification: diam(E) = diam(closure(E)) Since $E \subseteq cl(E) $, then it is immediate that diam $(E) \leq $ diam(cl($E))$.
I only need to show that assuming diam $(E) < $ diam(cl($E))$ will lead to contradiction then I can conclude that diam $(E)= $ diam(cl($E))$. 
So suppose that diam(cl($E$)) > diam($E$).     Then there exist a $p,q \in $ cl($E$)     such that $ d(p,q) > $diam($E$).  By def of cl($E$), there exist a sequence of $\{p_n\}, \{ q_n \}  \in E$ such that $ p_n , q_n  \to p, q ~~  $  respectively as $ n \to \infty $.  
Also note that by triangle inequality ,we have 
$$ (1) ~~~~~~ d(p,q) - [d(p,p_n) + d(q,q_n)] \leq d(p_n,q_n)   \text{  for all } n.$$
Since $ p_n, q_n \to p,q $ , we have $ d(p,p_n) + d(q,q_n) \to 0 $ as $ n \to \infty$.
Since $ d(p,q) > $ diam(cl($E$)), we can choose some $p_n, q_n $ such that 
$$  d(p,q) - [d(p,p_n) + d(q,q_n)] > \text{diam}(E).$$
Then by (1) above, we have 
$$ d(p_n,q_n) > \text{diam}(E),$$
a contradiction.  Thus the result holds.  
Is my proof correct? and is there a shorter way to do this?? thank you very much. 
 A: Suppose that $x,y\in\operatorname{cl}(E)$ and fix $\varepsilon>0$. Then, there exist $a,b\in E$ such that
\begin{align*}
d(a,x)<&\,\frac{\varepsilon}{2},\\
d(b,y)<&\,\frac{\varepsilon}{2}.
\end{align*}
The triangle inequality then implies that $$d(x,y)\leq d(x,a)+d(a,b)+d(b,y)\leq d(x,a)+\underbrace{\sup_{\hat a,\hat b\in E}d(\hat a,\hat b)}_{=\operatorname{diam}(E)}+d(b,y)\leq\operatorname{diam}(E)+\varepsilon.$$
Taking supremum over $x,y\in\operatorname{cl}(E)$, one has that $$\operatorname{diam}(\operatorname{cl}(E))=\sup_{x,y\in\operatorname{cl} E}d(x,y)\leq\operatorname{diam}(E)+\varepsilon.$$ Since $\varepsilon>0$ is arbitrary, it follows that $$\operatorname{diam}(\operatorname{cl}(E))\leq\operatorname{diam}(E).$$
A: Suppose diam(cl(E))>diam(E). Then diam(cl(E)-E)> 0. Therefore, for every p,q $\in$ cl(E) - E, $\exists$ d $\in \mathbb R\geq$ 0 such that d(p,q)< d.Now consider the following:By the definition of closure, cl(E)= $E\cup E'$ where E' = {p| p is an accumulation point of E}. Therefore, p,q $\in$ cl(E)- E are accumulation points of E where p,q$\notin$E.Clearly, there exists an open ball where $q\in B_d(p)\subset$cl(E) -E.  Let $N_p$(q) be a neihborhood of p containing q in cl(E)-E.Then since p is an accumulation point of E, there exists z$\in$E such that z$\in B_l$(p)$\subset N_p(q)$. where $B_l$(p)$\subset N_p$(q) where $l\leq d\in \mathbb R \geq 0 $.
Let r= diam(E). Then:
d(p,q) $\leq$ d(p,z) + d(z,q) = l + d = r.  
But this means diam(cl(E) $\leq$ diam(E) and we have a contradiction! Q.E.D.        
A: Suppose there exists $x, y \in \bar{E}$ such that $r = \rho(x, y) - \text{diam}(E) > 0$. Then there exists $u \in E \cap B(x, \frac{r}{2})$ and $v \in E \cap B(y, \frac{r}{2})$ (since $x, y \in \bar{E}$). Therefore
\begin{align*}
&& \rho(x, y) - \rho(u, v) \leq \rho(x, u) +   \rho(y, v) &< r = \rho(x, y) - \text{diam}(E) && \\
\Longrightarrow
&& \text{diam}(E) &< \rho(u, v), &&
\end{align*}
a clear contradiction. Thus, $\rho(x, y) \leq \text{diam}(E)$ for all $x, y \in \bar{E}$ and so $\text{diam}(\bar{E}) \leq \text{diam}(E)$.
