$\mathbb{Z}$ and $\{n + 1/n : n \ge 2\}$ are disjoint closed subsets of $\mathbb{R}$ at distance $0$, so the set-to-set distance is not a metric

No canonical integer lies strictly between $0$ and $1$: if $0 < k < 1$ in $\mathbb{R}$ for an integer $k$, then $0 < k < 1$ already in $\mathbb{Z}$, because the embedding is order preserving and injective and the order of $\mathbb{Z}$ is total; then $k > 0$ makes $k$ the image of a natural $n \ne 0$, so $n \ge 1$ and hence $k \ge 1$, contradicting $k < 1$.

$B$ is $\tfrac{1}{2}$-separated: for naturals $p > q \ge 2$ one has $p - q \ge 1$ and $0 < 1/p < 1/q \le 1/2$, so $(p + 1/p) - (q + 1/q) = (p - q) - (1/q - 1/p) \ge 1 - 1/q \ge 1 - 1/2 = 1/2$; hence distinct elements of $B$ differ by at least $1/2$ in absolute value.

A $c$-separated set $E \subseteq \mathbb{R}$ is closed: let $x \notin E$. The ball $B(x, c/2)$ contains at most one point of $E$, since two distinct points of it would be within $c/2 + c/2 = c$ of each other, contradicting $c$-separation with a strict inequality. If it contains none, then $B(x,c/2) \cap E = \emptyset$. If it contains exactly one point $e$, then $x \ne e$ gives $t := |x - e| > 0$, and the radius $r := \min\{c/2,\ t\} > 0$ has $B(x,r) \cap E = \emptyset$, because any point of $E$ in $B(x,r) \subseteq B(x,c/2)$ would have to be $e$, while $|x - e| = t \ge r$ puts $e$ outside $B(x,r)$. So the complement of $E$ is open and $E$ is closed.

$A$ is $1$-separated: for integers $m \ne m'$ the difference $k := m - m'$ is a nonzero integer, so by step 1.1 and trichotomy either $k \ge 1$ or $-k \ge 1$, and in both cases $|m - m'| = |k| \ge 1$.

$A$ and $B$ are disjoint and both nonempty: $0 \in A$ and $2 + 1/2 \in B$; and if $n + 1/n = m$ for a natural $n \ge 2$ and an integer $m$, then the integer $m - n$ satisfies $m - n = 1/n$ with $0 < 1/n < 1$ by [L2], contradicting step 1.1.

$A$ and $B$ are closed subsets of $\mathbb{R}$, by steps 2.1, 1.2 and 1.3 with $c = 1$ and $c = 1/2$ respectively.

$d(A,B) = 0$: the set $\{|a - b| : a \in A, b \in B\}$ is nonempty and bounded below by $0$, and for each natural $n \ge 2$ it contains $|n - (n + 1/n)| = 1/n$; given a real $\varepsilon > 0$, [L3] supplies a natural $n \ge 1$ with $1/n < \varepsilon$, and then $n + 1 \ge 2$ with $1/(n+1) < 1/n < \varepsilon = 0 + \varepsilon$, so the infimum is $0$ by the epsilon characterisation.

Yet every single point of $B$ is at positive distance from $A$: for $b \in B$ we have $b \notin A = \overline{A}$ by steps 2.2 and 3.1, so $d(b,A) \ne 0$ by the description of the closure, and $d(b,A) \ge 0$, hence $d(b,A) > 0$.

So $A$ and $B$ are nonempty disjoint closed subsets of $\mathbb{R}$ with $d(A,B) = 0$ and $A \ne B$: the separation axiom (M1) fails for the set-to-set distance, and it fails even on closed sets and even though each individual point-to-set distance is strictly positive.