We take a metric space $(E,d)$ and consider two closed subsets $A,B$ having a distance $d(A,B)$ equal to zero. We raise the following question: can $A$ and $B$ be disjoint – $A \cap B=\emptyset$?

If $A$ or $B$ is compact, let’s say $A$, $A \cap B$ cannot be empty. The proof is quite simple. As $d(A,B) = 0$, for all $n \in \mathbb{N}$ we can pick up $(a_n,b_n) \in A \times B$ with $d(a_n,b_n) \le \frac{1}{n+1}$. As $A$ is compact, we can find a subsequence of $(a_n)_{n \in \mathbb{N}}$ converging to a point $a \in A$. Even if it means renumbering the subsequence, we can suppose that $\lim\limits_{n \to +\infty} a_n = a$. As $d(a,b_n) \le d(a,a_n)+d(a_n,b_n)$ and both terms of the right hand side of the inequality converge to zero, $(b_n)_{n \in \mathbb{N}}$ also converges to $a$. As $B$ is supposed to be closed, we have $a \in B$ and finally $a \in A \cap B$.

So we have to turn ourselves to the case where $A$ and $B$ are not compact to find a counterexample. If $E = \mathbb{R}^n$ with $n \ge 1$ that means $A$ and $B$ unbounded.

An example in the real line

Take $A=\{a_n = n ; n \in \mathbb{N}\}$ and $B=\{b_m = m + \frac{1}{m+2}; m \in \mathbb{N}\}$. $A$ and $B$ are disjoint as if $n=m + \frac{1}{m+2}$ and $n=m$ we get the contradiction $\frac{1}{m+2}=0$ while if $n \neq m$ we get the contradiction $1 \le \vert n-m \vert = \frac{1}{m+2} <1$. However $d(A,B) = 0$ because $\lim\limits_{n \to +\infty} d(a_n-b_n) =0$.

An example in the real plane

Take $A=\{(x,y) ; y \ge 1/x > 0\}$ and $B=\{(x,0) ; x \ge 0\}$.
$A$ and $B$ are closed subsets of the real plane and their intersection is empty. The proof is left to the reader.