Let $D\subseteq \mathbb{R}^k$. Assume that $Pr_{u,v\leftarrow U_D}[|u-v| <B]>p$ for some $B,p$, where $|u-v|$ is the L1 distance of the vectors.
$S\subseteq D$ is obtained by sampling $n$ elements uniformly from $D$. Give a lower bound, as tight as possible, on $P\equiv Pr_{S\leftarrow U_D}[\exists u,v\in S:|u-v|<B]$. The bound can depend on the $k$, which should be thought of as a small constant, i.e., $k<<|D|$.
Specifically, the bound has to be tighter than the following trivial one: $P\ge 1-(1-p)^{n/2}$. This is obtained by dividing the $n$ points to $n/2$ pairs that are sampled independently. Then, $1-P$, the probability that $\forall u,v\in S:|u-v|\ge B$, is at least as low as the probability that none of these pairs will have distance lower than $B$. Since these events are independent, we get $1-P\le (1-p)^{n/2}\Rightarrow P\ge 1-(1-p)^{n/2}$.
Intuitively, since $S$ has $\frac{n^2}{2}=O(n^2)$ pairs of points, each corresponding with a possible "closeness" event, we would expect to get something like $P\ge 1-(1-p)^{O(n^2)}$, following a similar calculation to the above. However, without event independence this not easy to show (and possibly not true).
We are thus looking for a bound $C$ s.t. $P\ge C, C=1-(1-p)^{\omega (n)}$. Or to show that it does not exist. $C$ can be a function of the constant $k$.