Does there exist a set of intervals, no 5 of which share a point, such that the interval graph is non-planar?

Question

I'm having trouble proving this, but I assume I will need to make use of Euler's formula.

Answer 1

Such sets do exist. Here's an example: take the interval graph generated by $$ \{[4,5], [6,7], [8,9], [1,10], [2,11], [3, 12]\}. $$ The first three intervals are disjoint, and the last three intervals intersect all three of them, and each other. No point is contained in five or more intervals because it would need to be contained in two of the first three, which don't intersect.

The intersection graph they form is the following:

It is not planar because it contains $K_{3,3}$ not just as a subdivision but as a subgraph: each of the first three vertices (drawn around the outside of the diagram above) is adjacent to each of the last three (drawn in the middle).