Let's rewrite the problem statement in terms of graphs: in the graph the degree of each vertex is not more than four. Prove that all edges can be painted in two colors so that there are no single-color triangles.
Let our graph have $n$ vertices. We prove the statement of the problem by induction on the number of vertices.

Base case: if $n=1, 2$ everything is obvious
Assume the statement is true for any graph with $n \le k$ vertices. Consider a graph with $n=k+1$ vertices. Select the vertex $A$ arbitrarily. Remove this vertex from the graph as well as all edges from $A$. Thus we have a graph with $n=k$ vertices, that, by the induction hypothesis, can be painted as needed. Now paint vertex $A$ with its edges. Consider three cases.

I. The degree of each vertex $A$ is not more than 2. Then paint all edges from $A$ in different colors, which proves the problem statement.

II. The degree vertex $A$ is equal to 3, that is there is exactly 3 edges $AX, AY$ and $AZ$ (Fig. 42); the dotted line means it is not significant whether there is a corresponding edge. We know that the edges of $\Delta XYZ$ have different colors. Let, for instance, $XY$ be painted in the first color $YZ$ – the second color. Let's paint the edges as shown on Fig. 42: $AX$ and $AY$ – in the second color, and $AZ$ – first. By the induction hypothesis the statement is proved.

Fig. 42

III. The degree vertex $A$ is equal to 4, that is there are the following edges from $A$: $AP, AQ, AR$ and $AS$.

a) If the graph on the vertices $P, Q, R, S$ is complete, we have a separate connected component on the vertices $A, P, Q, R, S$. Wherein by the induction hypothesis it is a complete graph. Then let's paint its edges as shown on Fig. 43 and the statement is proved. Since all other vertices form a graph with less then $k$ vertices, by the induction hypothesis, it is painted as needed.

Fig. 43

b) If the graph on the vertices $P, Q, R, S$ is not complete, suppose there that there is no edge $QR$. Let's paint the edges $AP, AQ$ in a color, opposite to the edge $PQ$, and the edges $AR, AS$ – in a color, opposite to the edge $RS$. It is straightforward to check that there are no single-color triangles. Hence, by the induction, the problem statement is proved.