We need to prove that in a complete graph with n vertices and k colors, where the induced graph on each color is connected, there exists a multicolor triangle. Denote the colors by 1, 2, 3, …, k and recolor all edges that are in any of the colors 4, 5, …, k into color 3. The new graph satisfies the condition of connectivity on each color and if there is a multicolored triangle for it, then the same triangle in the initial graph will also be multicolored. Therefore, we can consider that k=3.
Suppose that the statement is not true for a graph G, as we can choose G to have a minimal number of vertices. It follows from the minimality of G that after removing any vertex, the new graph will not be connected by any of the colors, and let it be color 1. Denote by G1,G2,…,Gt the color connectivity components 1 after deleting vertex A. Since G is color 1 connected, there exist Ai∈Gi for which AAi is color 1. The segment A1A2 is not color 1 because G1 and G2 are different components of connectivity. Let it be of color 2. If A1B is a color 1 segment of G1, then the segment A2B cannot be of color 1 because G1 and G2 are different connectivity components; cannot be of color 3, because then A1A2B is a multicolor triangle and is therefore of color 2. Analogously, it is proved that all segments between the points of G1 and G2 are of color 2. We obtained that all segments between any two connectivity components are either color 2 or color 3.
Now consider segments AX and AY in colors 2 and 3, respectively (such segments exist, since G is connected in each of the colors). Without limitation, we have the following two cases:
1. X,Y∈G1, then one of the triangles AXA2 and AYA2 is multicolored.
2. X∈G1 and Y∈G2, then one of the triangles AXA2 and AYA1 is multicolored.
The resulting contradiction shows that for every graph with the given properties there exists a multicolored triangle. □