Let $A,B$ be two subsets of the given metric space $(X,d)$ such that $\overline{A}\cap B=\overline{B}\cap A=\emptyset$. Prove that there exist open sets $U\supseteq A$ and $V\supseteq B$ such that $U\cap V=\emptyset$.
It's obvious that for each $a\in A$, we have $r_a:=d(a,B)>0$. Similarly for each $b\in B$, $r_b:=d(b,A)>0$.
Now want to construct a cover for $A$ and $B$ with somethings just similar to $\{B_{\frac{r_a}2}(a)\}_{a\in A}$ and $\{B_{\frac{r_b}{2}}(b)\}_{b\in B}$. But I can't find neither a uniform radius, nor a way to make the covers separate.
I don't think $d(A,B)$ is necessarily zero. If it were $0$, we had such a uniform radius $\frac {d(A,B)}3$ !
The sets $U=\bigcup_{a\in A}B(a,\frac13r_a)$ and $V=\bigcup_{b\in B}B(b,\frac13r_b)$ are evidently open with $A\subseteq U$, $B\subseteq V$.
Also $U\cap V=\varnothing$.
This statement is proved below.
If $x\in U\cap V$ then $a\in A$ and $b\in B$ must exist with:
$$x\in B\left(a,\frac{1}{3}r_{a}\right)\cap B\left(b,\frac{1}{3}r_{b}\right)$$
Then:
$r_{a}\leq d\left(a,b\right)\leq d\left(a,x\right)+d\left(x,b\right)<\frac{1}{3}r_{a}+\frac{1}{3}r_{b}$
and also
$r_{b}\leq d\left(a,b\right)\leq d\left(a,x\right)+d\left(x,b\right)<\frac{1}{3}r_{a}+\frac{1}{3}r_{b}$
Summation gives:
$r_{a}+r_{b}<\frac{2}{3}r_{a}+\frac{2}{3}r_{b}$ wich is not true.
A contradiction has been found.
edit:
My choice for $\frac13r_a$ was somehow for safety. I see now that $\frac12r_a$ would have worked too.