let $X$ be a metric space and $V \subseteq X$ an open set which seperates two subset $A$ and $B$ of $X$. Then there is an open set $U \subseteq X$ which seperates $A$ and $B$ and such that $\overline{U} \subseteq V$.
I have already proof that there is an open set $U$ but I cannot prove that $\overline{U} \subseteq V$, can someone explain how we can achieve $\overline{U} \subseteq V$. Thanks in advance.
If $x \in U$ and $U$ is open, then in a metric space $(X,d)$ there is some $r>0$ such that $B_d(x,r) \subseteq U$ by definition of openness in a metric space.
Now $V=B_d(x, s)$ with $s=\frac{r}{2}>0$ is open and the triangle inequality implies $\overline{V} \subseteq \{y: d(x,y) \le s\}$ and the latter set is a subset of $U$ again.
So apply this idea to your given $V$ when $V$ is non-empty, if $V$ is empty, take $U=\emptyset$ instead.