Let $R$ be a rectangular region with sides $3$ and $4$. It is easy to show that for any $7$ points on $R$, there exists at least $2$ of them, namely $\{A,B\}$, with $d(A,B)\leq \sqrt{5}$. Just divide $R$ into six small rectangles with sides $2$ and $1$ and so at least one such rectangle must contain $2$ points from the seven ones. Thus the result follows.
Here is the question:
What about six points?
I believe that the same is true. How do I prove my belief?
ps: I don't want to find such $6$ points. I'd like to show it for any set with $6$ points.
The problem and solution are in Jiří Herman, Radan Kučera, Jaromír Šimša, Counting and Configurations: Problems in Combinatorics, Arithmetic, and Geometry, page 272. Let the rectangle have corners $(0,0),(0,3),(4,0),(4,3)$. Draw line segments joining $(0,2)$ to $(1,1)$ to $(2,2)$ to $(3,1)$ to $(4,2)$, also $(1,1)$ to $(1,0)$, $(2,2)$ to $(2,3)$, and $(3,1)$ to $(3,0)$. This splits the rectangle into $5$ pieces, and it's not hard to show two points in the same piece must be within $\sqrt5$.
A picture to illustrate the solution.