Hi, I don't think that the example of Gromov-Hausdorff is really an example. I mean, that is certainly a bounded metric space but is not compact. To see why consider the sequence where the $n$-th term is the space of $n$ points each with distance 1 from each other. This has no convergent subsequence in the sense of GH.
A set of metric spaces si relatively compact w.r.t. GH if and only if is it uniformly totally bounded, i.e. for every $\epsilon$ there exists $N(\epsilon)$ such that every space in the set can be covered with at most $N(\epsilon)$ balls of radius $\epsilon$ (Gromov).
P.s. I would have liked to post this as a comment, but don't really know how to, first post here
ADDED. Perhaps not really obvious, but a quite short argument gives the answer: what is true is that their distance is always at least $1/2$ ($1/2$ is realized by your example taking as second space the one with 2 points). Indeed assume by contradiction that for $n< m$ the distance between $X_n$ and $X_m$ is less that $1/2$. This means that there exists a metric space $(Y,d_Y)$ containing (isometric copies of) the spaces $X_n,X_m$ such that the Hausdorff distance of $X_n$ and $X_m$ in $Y$ is less than $1/2$.
Let $x_1,...,x_n$ be the points in $X_n$ and observe that by assumption for every $x_i\in X_n$ there exists a $y_i\in X_m$ such that $d_Y(x_i,y_i)<1/2$. Given that distinct points in $X_m$ have distance 1, such $y_i$ is unique for every $i$. But given that $m>n$ there must be $y\in X_m$ which is not in the set $(y_i)_{i=1,...,n}$. Then the triangle inequality tells that it must hold $d_Y(y,x_i)>1/2$ for every $i=1,...,n$ contradicting the assumption