= Solution
Fix $t$ and put $\sigma=t\wedge\tau$. The <stopping time> property implies that $\sigma$ is an $\mathcal F_t$-measurable <random variable> with values in $[0,t]$: for $s<t$,
$$
\{\sigma\leq s\}=\{\tau\leq s\}\in\mathcal F_s\subseteq\mathcal F_t,
$$
and for $s\geq t$ the event is the whole space. By part (c), the restriction of $X$ to $[0,t]\times\Omega$ is $\mathcal B([0,t])\otimes\mathcal F_t$-measurable.
The evaluation map $\omega\mapsto(\sigma(\omega),\omega)$ is measurable from $(\Omega,\mathcal F_t)$ to this product space: the inverse image of a measurable rectangle $A\times C$ is $\{\sigma\in A\}\cap C$. Composing it with the jointly measurable <stochastic process> gives an $\mathcal F_t$-measurable <random variable>
$$
X^\tau_t=X_{t\wedge\tau}.
$$
Since this holds for every fixed $t$, \b[the <stopped process> is adapted]. If path regularity holds only <almost surely>, first apply the proof to its pathwise regular representative; with a completed filtration, the original stopped variable differs only on a null event and is also $\mathcal F_t$-measurable. The <adaptedness of a stopped right-continuous process> requires neither boundedness of $\tau$ nor a <martingale> assumption; $\tau=\infty$ is harmless because $t\wedge\tau\leq t$.
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