But, I understand and trust the mapping of the problem to a graph, so perhaps if I see why the problem is hard on a graph that should be enough for me.

But my reasoning was along the lines (but now in my newfound terminology) of:

K1 - trivially 1 colour needed. K2 - 2.

If we add another region on the map, we either have K3 or K1,2. In the latter case the third region/vertex need only be different to the one it touches, but K3 clearly needs 3 colours.

Adding the fourth region is similar, and again K4 clearly needs 4.

The fifth region touches at most 3 others, since K5 is non-planar. Thus, it can take the colour of the untouched region.

That's not very formal, but why isn't proof by induction easy from there? i.e. if we can't build a map that ever requires a fifth colour, then it can't be that there exists a map which requires more than four?

* It's easy starting from the end, just put one colour in the middle, then alternate second/third colours around the edge until you get back to the start and have to add a fourth. From construction, the worst case is from K1,2 then adding another for K1,3 at which point we need four colours already, but when we add two more on what will be the C5# around they each only connect to one each side and the center. In adding each one you can simply use the colour one hop over on the circumference.

# Since the center of K1,3 must be the center of the final C5-plus-thing, as it already had 3 spokes.

It's just that the C₅-plus-thing requires four colors, even though it doesn't contain any K₄ induced subgraphs. So we could imagine that there might be some larger planar graph that requires 5 colors without containing K₅. As it turns out, there isn't, but that's the thing that's hard to prove.