Let $T$ be an optional time for the filtration $\{ \mathscr{F}_t \}_{t \ge 0}$, and define the sequence of random times $T_n$ by:
\begin{align*} T_n = T = \infty \; \text{on} \; \{T = \infty\}, \; \; T_n = \frac{k}{2^n} \; \text{on} \; \left\{ \frac{k-1}{2^n} \le T < \frac{k}{2^n} \right\} \\ \end{align*}
Show that $T_n$ are discrete stopping times for the discrete filtration $\{ \mathscr{F}_{\frac{k}{2^n}} \}_k$.
Considering the finite case first. For any finite $T$, integer $n \ge 1$, integer $k \ge 0$, we have:
\begin{align*} \left\{ \frac{k-1}{2^n} \le T < \frac{k}{2^n} \right\} = \left\{ T < \frac{k-1}{2^n} \right\}^c \cap \left\{ T < \frac{k}{2^n} \right\} \in \mathscr{F}_{\frac{k}{2^n}} \\ \left\{ T_n - \frac{1}{2^n} \le T < T_n \right\} = \left\{ T_n \le T + \frac{1}{2^n} \right\} \cap \{T_n \le T\}^c \in \mathscr{F}_{\frac{k}{2^n}} \\ \end{align*}
Since the above is for any $T$, can we just set $T = \frac{k}{2^n}$ to get $\{T_n \le \frac{k}{2^n}\} \in \mathscr{F}_{\frac{k}{2^n}}$ which would give the desired conclusion?
Am I close?