Problem showing that $\chi(G) \le 3$ constructing $G$ from a finite set of lines and maybe using greedy algorithm

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Consider a set of lines in the plane such that there are not three that intersect at the same point, and construct a graph $G$ whose vertices are the intersections of the lines, and where two vertices are adjacent if they appear consecutively on the same line. Show that $$\chi(G) \le 3$$

So I understand how $G$ is constructed, I had an idea from rotating the plane and coloring from left to right but at the end it didn't work.

Obviously the plane is in $\mathbb{R^2}$

I saw I maybe can do it using the greedy algorithm but using this give at most $Max degree + 1$ colors, how can I use it to only use 3 colors?

Is there really a way to color $G$ with at most 3 colors?