I can give a slightly better lower bound than v2/2. Again, let G1 be the grid Z^2 in R^2. Let G3 be G1 translated by (1/2, 0), then rotated 45° The two points of G3 nearest the origin O are A = (x, x) and B = (-x, -x), where x = v2/4. Consider a bijection f:G3 <=> G1. We must have either f(A) != O or f(B) != O. If f(A) != O, the closest possible point is f(A) = (1, 0) or f(A) = (0, 1). Either gives |A-f(A)| = d = v(5-2v2)/2 = .7368128+ By symmetry, if f(B) != O, |B -f(B)| = d as well. I claim that for the original G1 and G2, we can find a point P of G1 such that the closest points of G2 to P are arbitrarily near P+A and P+B. We have either f(P+A) = P or f(P+B) = P. If f(P+A) = P, then |P+B - f(P+B)| is arbitrarily close to d. Similarly If f(P+B) = P, then |P+A - f(P+A)| is arbitrarily close to d. This means d = v(5-2v2)/2 = .7368128+ is a lower bound of the required distance, slightly better than v2/2 = 0.7071067+