Prove correctness of DFA ending with ab
There is no induction needed. There is only one transition reaching the final state $q_2$, namely $q_1 \xrightarrow{b} q_2$. Furthermore, every transition reaching $q_1$ is labelled by $a$. It follows that every word accepted by the automaton has to end by $ab$. It remains to prove that, conversely, any word of the form $uab$ is accepted. After reading $u$ from the initial state $q_0$, you reach one of the three states $q_0$, $q_1$ or $q_2$. It suffices now to observe that $q_0 \xrightarrow{ab} q_2$, $q_1 \xrightarrow{ab} q_2$ and $q_2 \xrightarrow{ab} q_2$. Thus $q_0 \xrightarrow{uab} q_2$ in all cases, which means that $uab$ is accepted.