Let $a<b$ be real numbers. Suppose we have times $0=\tau_0<\sigma_1<\tau_1<\sigma_2<\tau_2<\cdots$ on some $\sigma$-finite filtered measure space $(X,\mathscr A,(\mathscr A_n)_{n\in\mathbb N_0},\mu)$, given by the equations $\sigma_k=\inf\{n>\tau_{k-1}:u_n\leq a\}, \tau_k=\inf\{n>\sigma_k:u_n\geq b\}$ for some adapted integrable sequence $(u_n)_{n\in\mathbb N_0}$. If we want to proof $\sigma_i,\tau_j,i\geq1,j\geq 0$ are stopping times, we could use induction. Then in the induction step, assuming $\tau_0,\sigma_1,\tau_1,\ldots,\sigma_{k-1},\tau_{k-1}$ are stopping times, we want to show that $\{\sigma_k\leq m\}\in\mathscr A_m$, or, looking at the complements, show that $\{\tau_{k-1}\leq m\}=\{\sigma_k> m\}\in\mathscr A_m$, which is clear since $\tau_{k-1}$ is a stopping time by the induction hypothesis.
To me, this last equation $\{\tau_{k-1}\leq m\}=\{\sigma_k> m\}$ is clear, looking at the definitions of the $\sigma_i,\tau_j$. However, if we again look at complements, we would get $\{\tau_{k-1}> m\}=\{\sigma_k\leq m\}$ which is nonsensible, since it imlies $m<\tau_{k-1}<\sigma_k\leq m$. I'm quite sure there is something I misunderstood; I just don't see what. Any help is much appreciated.
As you noticed $\{\tau_{k-1}\leq m\}\ne \{\sigma_k> m\}$ (this leads to an absurd conclusion). Instead, you may write $$ \{\sigma_k\le m\}=\bigcup_{i=1}^m\{\tau_{k-1}<i \}\cap \{X_i\le a\}, \quad k\ge 1, $$ and $$ \{\tau_k\le m\}=\bigcup_{i=1}^m\{\sigma_k<i \}\cap \{X_i\ge b\}. $$