Two color theorem

Proof: consider any country c on such a map. Now consider for any number n the set D(n) of countries at distance n from c. That is, a country is in D(n) iff it can be reached from c in n steps, but not in less. For n=0, the set consists of the country itself. For n>1, all neighbours of countries in D(n) are either in D(n-1) or D(n+1); if two countries in D(n) were neighbours, their corner point at the D(n-1) borderline would be surrounded by an odd number of countries! Therefore, we can colour D(n) in c's colour if n is even, and in the other colour if n is odd.

Attempts to extend this proof to the four color theorem have turned out to be nontrivial.

In fact, every map for which rp's condition holds is "topologically equivalent" to such a map. And we can prove the theorem in alternative manner for "my" maps.

For maps with 0 lines, the theorem is trivially true. Now suppose we want to add a line (straight or circle). We set a direction along the line, and flip the colours of all countries on its left. This transforms a legal 2-colouring of the map without the new line into a legal 2-colouring of the map with the new line. By induction, we see that such a map may be 2-coloured.