$n$ lines in a plane, proper coloring of intersection points with just 3 colors

Let us assume WLOG that no lines are vertical--if not rotate the plane to make this so [make sure you see that this is indeed possible].

Let us colour the set of points from left to right [no lines are vertical, thus for each line $L$ there is indeed a strict ordering of the points on $L$ from left to right], with the colours 1,2,or 3, as follows.

1. Colour a leftmost point with the color 1. [As there are only $<n^2$ points there is indeed a leftmost point. Pick one such point arbitrarily]

2. Let $p$ be a point that satisfies the following: Letting $L_1$ and $L_2$ be the two lines containing $p$ [as only two lines intersect a point there are only two such $L_i$], all points to the left of $p$ on $L_1$ [which recall neither $L_1$ nor $L_2$ is not vertical] have already been colored and all points to the left of $p$ on $L_2$ have already been colored. Then letting $p_i$ be the point immediately to the left of $p$ on $L_i$; $i=1,2$; and writing as $c(p_i)$ the color assigned to $p_i$ for $i=1,2$ [$c(p_i) \in \{1,2,3\}$], let the color $c(p)$ of $p$ be an integer in $\{1,2,3\}$ that is neither $c(p_1)$ nor $c(p_2)$. [There indeed always exists such a $p$ as long as there are still uncoloured points, as a leftmost point of the points not coloured yet always suffices.]

See for yourself that the above algorithm indeed terminates, and that it also terminates with a proper 3-colouring.