We can even assume the lattice is triangular: for each hexagon, add a point in its center and connect it to the six vertices of the hexagon. The conclusion holds for this larger lattice as well: Assume WLOG that we can find points $ A,B $ belonging to this lattice s.t. if $ O $ is the origin of the plane, $ OB $ is obtained from $ OA $ through a rotation of angle $ \frac\pi 2 $ around $ O $. This is equivalent to finding integers $ a,b,c,d $ s.t. $ \frac{a+b\varepsilon}{c+d\varepsilon}=i $, where $ \varepsilon $ is a solution to $ x^2+x+1=0 $ ($ a+b\varepsilon $ represents the point $ A $, while $ c+d\varepsilon $ - the point $ B $). In turn, this implies that we can find rationals $ u,v $ with $ u+v\varepsilon=i $. This is claerly absurd, since eliminating $ u $ between $ u+v\varepsilon=i,u+v\bar\varepsilon=-i $ gives an irrational value for $ v $.