Construct a set of points having some properties (colouring related) 
Construct a finite set of points $S$, all in the same plane, such
  that:
  
  
*
  
*Every line in the plane intersects $S$ in no more than $4$ points.
  
*If the points of $S$ are arbitrarily coloured with two colours, there are 3 collinear points of $S$ having same colour.
  


I think the following set $S$ of $13$ points would solve the problem (the horizontal lines are parallel):

Obviously the first condition is fulfilled.
Now, suppose there are two colours, blue and green. It seems that no matter how I distribute them, I end up with having 3 points with the same colour on the same line.
But I cannot give a rigorous proof and I need help with this (of course, listing all the possibilities is not an option).
 A: I'll number the points from top to bottom and on each line from left to right.  We will try to construct a coloring that fails to satisfy condition $2$ and show that we can't do it.  I'll think of the colors as red and green.
The corners of the triangle are points $1, 10$, and $13$.  If all three corners are red, then the inside points of the exterior edges, $2, 5, 6, 9, 11,$ and $12$, all must be green or one of the exterior edges satisfies condition $2$.  But if $2$ and $5$ are both green, then $3$ and $4$ must both be red, and if $6$ and $9$ are both green, then $7$ and $8$ must both be red, and that means, for example, that $1, 3,$ and $7$ are collinear red points.
Thus, both colors must be represented on the corners of the triangle.  Let's assume $1$ is red and $10$ and $13$ are green.  Then $11$ and $12$ must both be red, so $3, 4, 7,$ and $8$ must be green, and $2, 5, 6,$ and $9$ must be red, which means that $1, 2,$ and $6$ are collinear red points.
By symmetry, the last possibility is that $1$ and $10$ are red and $13$ is green.  Then $2$ and $6$ must be green.  We also know that $11$ and $12$ must be different colors.  First, assume $11$ is red.  Then $3$ and $7$ must be green, so $4$ and $8$ must be red and $1, 4$, and $8$ are collinear red points.  If $11$ is green and $12 $ is red, then $4$ and $8$ must be green so $3$ and $7$ must be red and $1, 3,$ and $7$ are collinear red points.
We have exhausted all possibilities so we have proved it is impossible to color the points without ending up with the collinear points of the same color.
A: Suppose there is a colouring scheme that contradicts the second requirement. Then on every 4-points-line one must have 2 green points and 2 blue points. 
Let's set a weight for each point: if the colour is blue the weight is 1 otherwise the weight is -1.
Now, let's calculate the weight's sum for all points in two ways:
1) Add all the weights on the same horizontal line. Then $$Sum = 1 + 0 + 0+0$$
2)  Add all the weights on the oblique lines (I'll consider the upper blue point on the first oblique line from the left). Then 
$$ Sum = 0 + (-1) + (-1) + (-1) = -3$$
The two different values for the sum provides the contradiction.
