The intent is not to give an example, but to give an algorithm. That is, to prove that an even plane triangulation is $3$-colorable, it gives a procedure for $3$-coloring such a plane triangulation:
- $2$-color the faces of the triangulation.
- Pick an arbitrary face and label its vertices $a$, $b$, $c$ either clockwise or counterclockwise, depending on the color.
- Pick an arbitrary face such that one of its vertices has already been colored. Complete its coloring using the color of that vertex and going either counterclockwise or clockwise depending on the color.
- Repeat step 3 until all vertices have been colored.
However, this is still not a complete proof because it's not immediately obvious that this algorithm doesn't lead to a contradiction somewhere along the way. Maybe we go around the graph in one direction and give a vertex label $a$, but then go around the graph in another direction and the algorithm tells us to give an adjacent vertex label $a$, too.
(It is worth noting that if the algorithm does work, then the output depends only on the way that step 2 is done; steps 3 and 4 are forced. So the order in which we consider faces is not relevant.)
One way I've seen of extending this argument into a complete proof is to generalize it. We call a planar embedding of a graph a near-triangulation if all faces, except possibly the unbounded face, are triangles. The more general claim is that all near-triangulations in which all vertices not adjacent to the unbounded face have even degree, can be given a $3$-coloring (of the vertices).
For such graphs, we can prove that this algorithm works by an induction on the number of bounded faces. Given a graph, we must:
- Delete an edge along the unbounded face, getting a graph with fewer bounded faces.
- Apply the algorithm to this graph; we know it works by the inductive step.
- Put the edge back in. Prove that its endpoints have different colors, and that the face it creates is colored in a way that follows the algorithm.