The inductive invariant is the correctness property itself (when all processes are done then at least one y is 1) in conjunction with "for all processes, when the process is not on line 1, then its x value is 1". You can probably replace the second conjunct with "for all processes, if the x value is 0, then the process is not done", but the proof gets a bit harder.
We need to show that Inv implies PartialCorrectness (trivial), that Inv holds in the initial state, and that if it holds in any state s, then it holds in any possible consecutive step s'. It's easy to see that it holds in the initial state. Now, let's assume it holds in s, and prove for s'. To make this transition, some process p either executes line 1 or executes line 2. If it executes 1, then PartialCorrectness doesn't change because no new process is done. The second conjunct holds because we've just left line 1 and x has been assigned (by the definition of line 1). If we are currently in line 2, the second conjunct of the invariant doesn't change. By the definition of this action, we'll be done. Here we have two cases, either we set y to 1, or we set y to zero. If we set y to 1, we're done and PartialCorrectness holds. If we set y to 0 then by the assumption of the invariant, the process we depend on must not be done, hence AllDone is false, and PartialCorrectness holds. QED.