Example for $A,B$ non-empty, bounded, and closed, $A\cap B = \emptyset$, but $\operatorname{dist}(A,B) = 0$ Is there such an example in a Banach space? Note that this is not possible in a finite-dimensional space.
In what I have here, $A$ and $B$ are the closures of relatively open sets in a bounded and closed set.
 A: Inspired by this:
Consider the Banach space $\ell^2$. Define $A$ to be the set of sequences for which only one term is $1$. This is bounded and non-empty and closed.
Define $B$ to be the set of sequences of the form
$$ \mathbf b = (0,\dots,0,1+2^{-n},0,\dots)
$$
where $n$ is the index of that non-zero term in the sequence. This is also bounded (all terms are within 2 distance from 0) and non-empty and closed.
$A,B$ are disjoint.
Then for all $\delta > 0$ we can find $\mathbf a \in A$ and $\mathbf b \in B$ with $\lVert \mathbf a - \mathbf b\rVert < \delta$. Indeed, take $\mathbf a$ to be the sequence for which $a_n = 1$ and $\mathbf b$ to be the sequence for which $b_n = 1 +2^{-n}$ for the same $n$. For sufficiently large $n$ $1 + 2^{-n} - 1$ can be arbitrarily small. Hence $\operatorname{dist}(A,B) = 0$.
A: Note that the distance is $0$ iff $0$ is in $\overline{A-B}$. 
But it is known that the sum of two closed subsets need not to be closed. See this. (In the link, we find two subspaces $U,V$ of $X$ such that $U+V\ne X$ but $\overline{U+V}=X$. To get a bounded example, we take unit balls $A,B$ of $U,V$. Then $A+B$ is not closed. This is obvious in that example, but I am not sure if this holds in general.)
Hence, you can take such $A,B$ and some point $x\in \overline{A-B}\setminus A-B$. You can assume $x=0$. This will provide an example you want.
