It does appear to be a proof by contradiction, at least for observables. If A is not Hermitean then \langle a|A=(A^\dagger|a\rangle)^\dagger\neq (A|a\rangle)^\dagger, so acting A to the left in the term \langle a|AB|a\rangle doesn't yield a\langle a|B|a\rangle, as required to obtain the...