Let $G$ be a graph with 9 vertices whose complement, $\overline{G}$, has three components.
I've read that it can't be planar, but am having trouble completing a proof.
Most ways of distributing the 9 vertices among the 3 components of $\overline{G}$ would force $G$ to be non-planar, it is clear to me. For example, if 6 vertices were in one component, 2 in another and 1 in the last, then in $G$ any 3 vertices from that 6-vertex component would be completely connected to the vertices from the other 2 components, i.e. $K_{3,3}$ would be a subgraph of $G$, so $G$ would be non-planar.
This argument works when the largest component has 5 vertices, 4 vertices, or even 3 vertices.
That leaves one case: when 7 vertices are connected into a component in $\overline{G}$ and the other 2 vertices are isolated points. Why must $G$ be non-planar in this case?
In fact, the statement is wrong: $G$ can be planar.
Assume that the component of size 7 in $\overline{G}$ is the clique; then $G$ is the tripartite graph $K_{1,1,7}$, which is planar: