As the sibling comment to this says, you must still be confused about something here. Let me expand. The proof that every planar graph (equivalently every map

As the sibling comment to this says, you must still be confused about something here. Let me expand. The proof that every planar graph (equivalently every map on a plane) is five colourable proceeds as follows. Cleverly produce a set U of small graphs S that have the following properties: 1: For every graph G, there is an S in U such that S is a sub-graph of G; 2: If G is a counter-example to the theorem, and S from U is a sub-graph of G, then there is a smaller graph H that is also a counter-example. This is enough, because now let M be a minimal counter-example. By (1) there is an S from U that is a sub-graph of M. But by (2) there is then a smaller counter-example, contradicting the minimality of M. So there can be no counter examples. With that now in mind, here is the original statement: We proceed by showing that there is a small set of sub-graphs such that every graph must contain one (an unavoidable set), and then showing for each of those that it cannot be in a minimal counter example. This is easily done by hand. I hope that helps.