$d(A,B)=\min\{ d(a,b): a\in A\text{ and }b\in B\}>0$ then $A\cup B$ is not connected 
If $d(A,B)=\min\{ d(a,b): a\in A\text{ and }b\in B\}>0$ then $A\cup B$ is not connected.

We claim that the two sets are disjoint, suppose it is not we have $d(A,B)=0$
So now we have to say any union of two arbitrary disjoint sets is not connected
I am stuck here
 A: As A and B are disjoint then you can show that They are actually both open and close in AUB considering the relative topology. Hence they will form a separation of A and B.
A: Union of two disjoint sets can be connected: $(0,1)$ and $\{0\}$ are disjoint but their union is connected. 
Let $C=A \cup B$. If $C_1=\overline {A} \cap C$ and $C_2=\overline {B} \cap C$ then $C=C_1\cup C_2$ and $C_1,C_2$ are non-empty disjoint closed subsets of $C$, so $C$ is not connected. [They are disjoint because $d(\overline {A} ,\overline {B} )=d(A,B)>0$]. 
A: So there exists $r>0$ with $d(a,b)>2r$ for all $a\in A$, $b\in B$.
What can you say about the following unions of open balls of radius $r$?
$$\bigcup_{a\in A}B_r(a)\qquad\text{and}\qquad \bigcup_{b\in B}B_r(b)$$
A: *

*In general,  if $d(A,B)>0$ then $A$ and $B$ are disjoint sets.

*If $d(A,B)=0$ then $A$ and $B$ may be or may not be  disjoint sets 
A: Note that for the closure of a subset $C$ of the metric space $X$ we have 
$\bar{C} = \{ x \in X \ | \ d(x,C)=0\}$
Back to our problem. Since $d(A,B)>0$ we have $d(a, B)>0$ for any $a\in A$ and so $A \cap \bar{B} = \emptyset$. Similarly, $\bar A \cap B = \emptyset$. 
Now, the closure of $A$ as a subspace of $A\cap B$ equals $\bar A \cap(A\cup B) = (\bar A\cap  A) \cup (\bar A\cap B)= A$, and so $A$ is a closed subspace of $A\cup B$. Similarly $B$ is a closed subspace of $A\cup B$. Hence, $A$, $B$ form a separation of $A\cup B$, and so $A\cup B$ is not connected.
