First of all, note that a graph which contains a non-planar subgraph obviously cannot be planar: adding new vertices and edges to an already non-planar graph can't possibly make drawing it in a plane without intersections any easier.
In your case, $L_i$ is a subgraph of $L_j$ for all $j > i$; thus, if we can show that $L_i$ is non-planar for some $i$, we've also shown it for all $j > i$. Conversely, showing that $L_i$ is planar implies that $L_j$ is also planar for all $j < i$.
It should not be particularly hard to draw the graphs from $L_1$ up to $L_4$ on paper without intersections, thereby demonstrating that they're planar.
As for $L_5$, recall Wagner's theorem: A finite graph is planar if and only if it does not have $K_5$ (the complete graph on five vertices) or $K_{3,3}$ (the complete bipartite graph on six vertices) as a minor. Can you find either of these graphs as a minor in $L_5$? (Hint: try contracting edges of the form $(l_i, r_i)$.)