Proof about convex polyhedron This exercise is from The Kürschák Mathematics Competition from 1947:

Prove that exept for any tetrahedron, no convex polyhedron has the property that every two vertices are connected by an edge. Degenerate polyhedron are not considered.

Every convex polyhedron can be represented as a $3$-connected simple planar graph (proof here). ($3$ connected here means that there does not exist any set of $k-1$ vertices whose removal disconnects the graph). Now our hypothesis says that we need to have a full graph (that is all vertices are connected to every other via an edge). And the smallest such is $K_4$ which is the polyhedra (polyhedral graph) of the tetrahedron and therefore the result follows.
First of all, is my reasoning correct?
Secondly, this was an exercise for high-school students, probably without knowledge of these results. So I would like to get some HINTS and IDEAS about how to show this with other methods (both high-school level or university level, anything is welcome)
Thank you!
 A: You can actually generalize it to say that a polytopes that has the corresponding property must be a simplex, but for a more down-to-earth proof you probably need to skip that generalization. 
A strictly geometric way to prove it is to note that given such a polytope you can construct a polytope with the corresponding property in one dimension less with one vertex less by consider a cone formed from one of the vertices and consider it's intersection with a hyperplane.
In short this means that a complete pentagonal polyhedron would guarantee the existence of a complete quadrigonal polygon and so on until we realize the absurdity.
A more direct way in the same theme would be to just consider a complete pentagonal polyhedron and realize that four of the vertices must lie in a plane in order to have the "diagonals" lying on the boundary of the polyhedron. But then they're a proper diagonal and not an edge of the polyhedron.

As for your reasoning it seems correct, but I don't see how that proves that it excludes polyhedra with more than $4$ vertices. 
