How to prove K3,3 has 2 as its thickness? The thickness of a simple graph G is the smallest number of planar subgraphs of G that have G as their union.
Show that K3,3 has 2 as its thickness.
My attempt: 
[If we decompose K(3,3) into any pair of nonempty graphs A and B, the union of their edges make the edges of K(3,3), and A and B have no edges in
common), then both A and B are planar. (Example: let A be one
arbitrary edge of K(3,3) and let B be the rest.) It follows that
the thickness of K(3,3) is no more than 2.]
It is enough? 
how to prove it more complete

This proves that thickness $\tau(K_{3,3})$ is at most $2$. To show that $\tau (K_{3,3}) >1$, note that $K_{3,3}$ itself is certainly not planar, so it's not possible to represent it as a union of 1 planar subgraph.
Thus, $2 \geq \tau (K_{3,3}) >1$, which implies $\tau (K_{3,3}) = 2$, since $\tau (K_{3,3})$ is an integer.
