Steinitz's theorem

Remarks

One direction is easier. The graph of a $3$-polytope is planar: project its boundary onto one face from a point just outside that face. It is $3$-connected by Balinski's theorem. For the converse, one proof reduces every $3$-connected planar graph to the complete graph $K_4$. It uses moves that replace a triangle by a three-legged star and back, and shows that each move can be carried out on polytopes. Another proof uses the circle packing theorem to realize the graph with all edges tangent to the unit sphere.

Example: the cube graph, with $8$ vertices of degree $3$, is planar and $3$-connected, and the cube realizes it. Without $3$-connectivity the conclusion fails: the cycle graph $C_4$ is planar but only $2$-connected. No $3$-polytope has it as its graph, because every vertex of a $3$-polytope has degree at least $3$.

Going deeper

For a student

The theorem lets one count and list combinatorial types of $3$-polytopes by counting $3$-connected planar graphs. It also shows that every combinatorial $3$-polytope can be realized with rational, even integer, coordinates, and with many extra properties, such as all edges tangent to a sphere. In dimension $4$ and higher no such characterisation is known, and realizability of a given face lattice is a hard problem.