When backward induction (BI) is well-defined (and it looks like it will be here), then it will result in the unique sub-game perfect equilibrium. To do it, start from the bottom of the tree that you have drawn and see what the working class player would do at each of those last four nodes. Anticipating that decision, what would the elite player prefer at each of the nodes above? Keep working upwards through the tree, anticipating the decisions below the node your'e looking at. If you ever get a tie, i.e. a player is indifferent between two or more options, then, technically speaking, BI is not well-defined. If you don't, however, you're guaranteed to get the subgame perfect equilibrium, and it will be unique. You'll have to do this for each case of $q$, as you describe.
There is the slight complication here in that, in the higher nodes, the players will be comparing expected payoffs across their options, but that does not stop you from using backward induction.
Let me know if you need any help applying BI here. I haven't actually done it myself yet, but it looks pretty clear to me that that is how the problem was intended to be solved.
Edit: For some specific $q$, there will be multiple subgame perfect equilibria. Formally speaking, you should use the method as you describe, converting it to normal form, finding NE and then eliminating those that aren't subgame perfect. On a more intuitive level, though, you can just find the subgame perfect equilibrium around those important thresholds (for instance, for $q=1/3$, check $q=1/4$ and $q=1/2$), and then argue that at the threshold, the subgame perfect equilibrium on either side of the threshold is a subgame perfect equilibrium at the threshold (i.e. the subgame perfect equilibria at $q=1/3$ will probably be the subgame perfect eqm. at $q=1/4$ and the subgame perfect eqm. at $q=1/3$). Again, to show that formally, you'd technically want to use the method you describe because BI is not well-defined, but that's how you can sort of intuit the answer using BI, which is a much easier algorithm to apply than the method you describe.