In Durrett's probability theory and examples, it asks suppose $M \le N$ are stopping times. If $A \in\mathcal{F}_M$, then $$L = \begin{cases} M & \text{on }\; A\\ N & \text{on} \; A^c \end{cases}$$ is a stopping time.
To show that $L$ is a stopping time, I believe what I need to show is that $ \{L= n \}$ is $\mathcal{F}_n$ measurable. I'm not sure how we can write $\{ L = n\}$ in terms of $M$ and $N$. What I have is $$\{ L = n \} = ( \{ M = n\} \cap A) \cup ( \{ N = n \} \cap A^c)$$ Since RHS is $\mathcal{F}_n$ measurable, we see that $\{ L = n\}$ is $\mathcal{F}_n$ measurable.
However, the textbook gives us a completely different proof where it states $$\{ N = n\} = (\{L = n\} \cap A) \cup \bigcup_{m=1}^n (\{ L = m\} \cap \{ M = n\} \cap A^c)$$ I can see how we can write $\{ N = n\} = (\{ L = n \} \cap A) \cup (\{ L = n\} \cap A^c)$ but how do we get $\bigcup_{m=1}^n (\{ L = m\} \cap \{ M = n\} \cap A^c)$?
Your proof assumes that $\mathcal F_M \subseteq \mathcal F_N$ which gives $A \in \mathcal F_N$ and hence $(N=n)\cap A^{c} \in \mathcal F_n$. The proof in the book does not assume this result and it uses only the definition of stopping times.