= Solution
{c}
For $A\in\mathcal F_S$, write
$$
A\cap\{S\leq T\}\cap\{T\leq n\}
=\bigcup_{k=0}^n\left[A\cap\{S=k\}\cap\{k\leq T\leq n\}\right].
$$
The first two factors together belong to $\mathcal F_k$, since $A\cap\{S=k\}$ is the difference of its intersections with $\{S\leq k\}$ and $\{S\leq k-1\}$. Also $\{k\leq T\leq n\}\in\mathcal F_n$ by the <stopping time> property. The filtration inclusion puts every term, and therefore the union, in $\mathcal F_n$. By definition,
$$
\boxed{A\cap\{S\leq T\}\in\mathcal F_T.}
$$
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