Drawing a complete graph with four vertices or less such that no edges cross is trivial. I conjecture, and would like to prove, that it is impossible with five. This is what I've come up with:

*The only way to draw a complete graph with three vertices $A$, $B$ and $C$ is to draw a triangle $ABC$ (fig. 1).*

[![Fig 1: Triangle ABC](https://i.stack.imgur.com/6K11p.png)](https://i.stack.imgur.com/6K11p.png)

Fig 1: Triangle $ABC$

*The only way to draw a complete graph with four vertices is to add a vertex $D$ inside or outside of $ABC$ (fig. 2).*

[![Fig 2: possible positionings of D](https://i.stack.imgur.com/FtUQX.png)](https://i.stack.imgur.com/FtUQX.png)

Fig 2: possible positionings of $D$

*In any case, we can rename the vertices, which results in an outer triangle $EFG$ containing $H$, split into three smaller triangles $EFH$, $FGH$ and $GEH$.*

*To create a complete graph with five vertices, we must add a vertex $I$ inside $EFH$, $FGH$ or $GEH$, or outside of $EFG$. In all cases, the edge linking $I$ to the vertex not in the aforementioned triangles (respectively $G$, $E$, $F$ and $H$) must cross at least one other.*

This "proof" feels unnecessarily long-winded, and maybe not very rigorous. Can anyone suggest a more efficient one?

Furthermore, this is the two-dimensional case. It seems obvious to me that in more dimensions, as long as you avoid having five vertices on the same plane, the edges should never need to cross; but is there a way to prove that?