Consider a graph $G$ with $100$ vertices, each represents a real number. We connect two of them if and only if their difference is exactly one. Clearly, the degree of each vertex is at most two. So the graph must consist of some pairwise disjoint path(s) and cycle(s). Moreover, we can easily prove that the cycle is impossible. Then we list all vertices in all path(s) consecutively $V_1, V_2, ..., V_{100}$ and put them so that $V_iV_{i+1}$ don't have any common side (this is obvious).