 Solution

In the final round, player 1 strictly accepts every $q_2>0$. In any subgame perfect equilibrium it must also accept $q_2=0$ with probability one. Otherwise, if that acceptance probability were $a<1$, offering zero would give player 2 only $a\alpha$. A sufficiently small positive offer would be accepted and yield more. No positive offer maximizes player 2's payoff either, because a smaller positive offer improves it. The same supremum argument rules out a mixed offer distribution attaining the missing maximum. Thus the final-round equilibrium uniquely has $q_2=0$, accepted, with payoffs $(0,\alpha )$.

In round one, player 2 therefore rejects $q_1<\alpha$ and accepts $q_1>\alpha$. At $q_1=\alpha$ it is indifferent, but equilibrium again forces acceptance with probability one: if acceptance were less likely, player 1 could improve on any proposed optimum with an offer just above $\alpha$. Since $\alpha <1$, player 1 strictly prefers the resulting positive payoff to rejection.

Consequently the unique equilibrium outcome is immediate agreement at $q_1=\alpha$, giving payoffs $(1-\alpha ,\alpha )$. The full strategies are: player 1 offers $\alpha$ initially and accepts every nonnegative final offer; player 2 accepts initial offers at least $\alpha$, rejects smaller ones, and offers zero in every reached final-round subgame. These prescriptions cover off-path histories too. Acceptance at indifference follows from the continuous-offer bargaining acceptance lemma, rather than an arbitrary tie-breaking convention.