Proof that a Octahedron graph is 3-colorable?

Question

I worked on a problem that gives an adjacency matrix and lets you draw a graph from that. After some time I found that the graph is a) planar and b) not any planar graph, but a Octahedron.

The next part asked for a coloring of the graph and a proof thereof. We know that every planar graph can be colored in at most 4 colors by the 4-Color-Theorem, but after some trying I found out the above graph can be actually colored in 3 colors. Is there any straightforward proof of that, besides a proof by picture that just shows the actual coloring?

It has to do with the triangles that exist in the graph, so I thought a bit about using some kind of structural induction that starts with one triangle and then adds more until the structure is complete, but I think that is also quite involved.

Answer 1

Each vertex is adjacent to four others, and antipodal to one. There are six vertices, hence three nonadjacent, antipodal pairs. Each antipodal pair receives one color.

Answer 2

If you know that the regular octahedron is the dual of the cube, then you know that $3$-coloring the vertices of the octahedron is equivalent to $3$-coloring the faces of the cube. If the cube is in standard position in a coordinate system, then each face of the cube is parallel to one of the three coordinate planes, and of course parallel faces can be given the same color.