= Solution
A discrete-time <martingale> relative to a <filtration> $(\mathcal F_n)$ is an <adapted process> $(M_n)$ with $\mathbb E|M_n|<\infty$ and
$$
\mathbb E[M_{n+1}\mid\mathcal F_n]=M_n\quad\text{almost surely}.
$$
Iterating <conditional expectation> also gives $\mathbb E[M_m\mid\mathcal F_n]=M_n$ for $m\geq n$. It is bounded in $L^2$ when each $M_n$ is square integrable and
$$
\boxed{\sup_{n\geq0}\mathbb E|M_n|^2<\infty.}
$$
This is a uniform bound on the second moments, not a pathwise bound on all sample values.
The <optional stopping theorem> for bounded <stopping times> says that, if $\sigma\leq\tau\leq K$ for a deterministic finite $K$, then
$$
\boxed{\mathbb E[M_\tau\mid\mathcal F_\sigma]=M_\sigma,\qquad
\mathbb EM_\tau=\mathbb EM_0.}
$$
Here $\mathcal F_\sigma$ is the <stopping-time sigma-algebra>. The second assertion follows by taking $\sigma=0$ and then <expectations>. An $L^2$ bound is not needed: integrability at the finitely many deterministic times suffices.
For a proof, both stopped values are integrable since $|M_\tau|,|M_\sigma|\leq\sum_{j=0}^K|M_j|$. Expand
$$
M_\tau-M_\sigma=\sum_{k=0}^{K-1}\mathbf1_{\{\sigma\leq k<\tau\}}(M_{k+1}-M_k).
$$
If $A\in\mathcal F_\sigma$, then $A\cap\{\sigma\leq k\}\in\mathcal F_k$ by the definition of the <stopping-time sigma-algebra>, and $\{\tau>k\}\in\mathcal F_k$ because $\tau$ is a <stopping time>. Thus the indicator of $A\cap\{\sigma\leq k<\tau\}$ is $\mathcal F_k$-measurable. Multiplying the sum by $\mathbf1_A$, taking <expectations>, and conditioning each summand on $\mathcal F_k$ gives zero, by the <martingale> increment property. Therefore $\mathbb E[\mathbf1_AM_\tau]=\mathbb E[\mathbf1_AM_\sigma]$ for every $A\in\mathcal F_\sigma$. Since $M_\sigma$ is $\mathcal F_\sigma$-measurable, this is exactly the claimed <conditional expectation> identity.
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