Don't understand closure under string reversal
Both $q$ and $r$ are states in this construction. If in original $\delta'$ we can step $r\stackrel{t}{\rightarrow} q$ then in the new $\delta''$  we can step $q\stackrel{t}{\rightarrow} r$. Like stated in the proof the arrows are reversed. 

You are right, $\Sigma_\epsilon$ is $\Sigma$ plus the empty word $\epsilon$. The original automaton initially has no $\epsilon$-transitions. However it has multiple final states, and after reversing it would have multiple initial states. A possible solution is given here, using $\epsilon$-transitions, connecting a new final state from the original final ones. This is the first step, from $\delta$ to $\delta'$.

If you are still trying to decipher the construction then of course the induction is not trivial. Once you see the arrows are just reversed, then it is not hard to come up with a possible induction assumption: if I can go from $q$ via string $w$ to $r$ in $\delta'$, then I can go from $r$ via string $w^R$ to $q$ in $\delta''$. Then it is no longer difficult. It would not say "easy" as that depends on your experience with induction.