Color with color one all rational points $(x,y)$ with $\text{min}(v_2(x),v_2(y))>0$, with color two all those points with $v_2(x)\ge v_2(y)$ and $v_2(y)\le 0$, and with color three all those with $v_2(x)<v_2(y)$ and $v_2(x)\le 0.$

Considering point of the form $(2(1+2k)/q,2(1+2l)/q)$ with $k,l\in\mathbb{Z}$ and $q$ odd we obviously have a point of color one within $2/q$ distance, both $x$-wise and $y$-wise, from any point in the plane. Letting $q$ tend to infinity, and arguing in a similar manner for the other colors, it's clear that $1)$ is satisfied.

Suppose that we have some points $(x_1,y_1)=(2^a\cdot p_1/q_1, 2^b\cdot p_2/q_2)$ (with $a,b>0$ and $p_i,q_i$ odd, hence of color one),$(x_2,y_2)= (2^c\cdot p_3/q_3, 2^d\cdot p_4/q_4)$(with $c\ge d$,$d\le 0$ and $p_i,q_i$ odd, hence of color two) and $(x_3,y_3)=(2^e\cdot p_5/q_5, 2^f\cdot p_6/q_6)$(with $e<f$, $e\le 0$ and $p_i,q_i$ odd, hence of color three) of different colors that are collinear. Then $$\begin{vmatrix} x_1 & y_1 & 1 \\ x_1 & y_1 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}=0.$$Rationalizing and multiplying with $2^{-d}$ the first row and by $2^{-e}$ the second row, we obtain $$\begin{vmatrix} 2^a\cdot\text{odd} & 2^b\cdot\text{odd} & \text{odd} \\ 2^{c-d}\cdot\text{odd} & \text{odd} & 2^{-d}\cdot\text{odd} \\ \text{odd} & 2^{f-e}\cdot\text{odd} & 2^{-e}\cdot\text{odd} \end{vmatrix}=0.$$Mod $2$ this cleans out as $$\begin{vmatrix} 0 & 0 & 1\\ x & 1 & y \\ 1 & 0 & z \end{vmatrix}=0,$$which by expansion along the first row becomes $$\begin{vmatrix} x & 1 \\ 1 & 0 \end{vmatrix}=0.$$But this last determinant is equal to $-1\neq 0$, a contradiction!