Proving greedy choice property of fractional knapsack
Remember that $O$ is assumed to be an optimal solution.
If $O'$ is a solution that is at least as good as $O$, then $O'$ has to be optimal too.

In particular, since $O$ is optimal, $O'$ cannot be better than $O$; so $O'$ must be exactly as good as $O$.  Since there's nothing better than $O$, and since $O$ and $O'$ have the same goodness, there can't be anything better than $O'$ either.  Consequently, $O'$ is also optimal: both $O$ and $O'$ are optimal solutions.  (There can be more than one optimal solution.)