Given a plane with lines intersecting each other, and there is no limit on how many lines can intersect at one point. We make graph $G$ from such line arrangement. Prove that $\chi(G)\leq4$ without using 4CT.
Apparently $G$ is planar, so $3\leq\chi(G)$, how should we prove the upper bound?
I found a case where 4 colors are needed: draw a $K_5$ graph, and turn intersections into vertices, and then we greedy color it from left to right (following the x axis from low to high value), then the right most vertex has color 4.