The problem is given as follow:
Let G be a connected graph having an even number of edges such that all the degrees are even. Prove that the edges of G can be colored by red and blue in such a way that every vertex has the same number of red and blue edges adjacent to it.
My approach is:
Since all degrees are even, then G is not a tree and contains only cycles. This also implies two cycles cannot share a common edge, otherwise one vertex has odd degree. G has even number edges indicates it has even number of cycles which share common vertices. Therefore, G can be colored by red and blue in such a way that every vertex has the same number of red and blue edges adjacent to it.
I don't know if I miss something needs to be clarify in this proof...
My new approach:
G contains a Eulerian circuit because all the degrees are even. Traversing the circuit from a vertex u and coloring the edges blue, red, blue, red....as a sequence. Because G has even number of edges, it has even number of blue and red edges; hence, the color of first edge starting from vertex u is different from the last edge goes back to starting point. Therefore, QED