Property of distance and adherence Please how to prove that in a metric space $(E,d),$ for $A,B\subseteq E$ that 
$\forall x\in E, d(x,A)=d(x,\overline{A})$ and that $d(A,B)=d(A,\overline{B})=d(\overline{A},\overline{B})$
and  $$\overline{A}=\overline{B}\Longleftrightarrow \forall x\in E, d(x,A)= d(x,B).$$
For the fist I know that $A\subset \overline{A}$ so $d(x,\overline{A})\leq d(x,A)$.
I think that $d(A,B)=\inf_{A} d(a,B)=\inf_{A} d(a,\overline{B})=d(A,\overline{B})$ (using the first equality).
But I don't know how to continue?
Thank you.
 A: $d(A,B)\ge d(\overline A, \overline B)$ true trivially, for the other direction let $\varepsilon>0$, then there are $x\in\overline A$, $y\in\overline B$ such that $d(x,y)\le d(\overline A, \overline B)+\varepsilon$. There are also $a\in A$, $b\in B$ such that $d(x,a)\le\varepsilon$ and $d(y,b)\le\varepsilon$. Using the triangle inequality we have
$$d(A,B)\le d(a,b)\le d(a,x)+d(x,y)+d(y,a)\le\varepsilon+(d(\overline A, \overline B)+\varepsilon)+\varepsilon=d(\overline A,\overline B)+3\varepsilon$$
The inequality follows by letting $\varepsilon\to 0$.
$d(x,A)=d(x,\overline A)$ is just a special case of this for $B=\{x\}$.
If $\overline A=\overline B$, then for any $x\in E$ we have $d(x,A)=d(x,\overline A)=d(x,\overline B)=d(x,B)$. For the other direction, let $x\in\overline A$, then $0=d(x,\overline A)=d(x,A)=d(x,B)=d(x,\overline B)$, so $x\in\overline B$, therefore $\overline A\subseteq\overline B$; the other inclusion is analogical.
A: Notice that for any $\varepsilon > 0$ there exists $a\in A$ such that
$$
 d(x,\bar A) \ge d(x,a) - \varepsilon.
$$
