Since $T$ is a stopping time, by definition, for any $t \ge 0$, $\{T \le t\} \in \mathscr{F}_t$.
$T \land t = \min(T, t)$
We have to show that $T \land t$ is $\mathscr{F}_t$ measurable, which means for any Borel set $B$, that $\{T \land t \in B\} \in \mathscr{F}_t$.
My attempt:
\begin{align*} (T \land t) \in B = (\{T \in B\} \cap \{T \le t\}) \cup (\{t \in B\} \cap \{T \ge t\}) \\ \end{align*}
On the right, the second, third, and fourth terms are clearly $\mathscr{F}_t$ measurable, but the first term, $\{T \in B\}$ isn't necessarily so. I'm stuck on what to try from there.
It is enough to show that $(T\wedge t \leq a)\in \mathcal F_t$ for ech real number $a$. $$(T\wedge t \leq a)=(T\leq t, T\leq a)\cup (T>t, t \leq a).$$ Note that $$(T\leq t, T\leq a)=T \leq t\wedge a)\in \mathcal F_{t\wedge a} \subseteq \mathcal F_t. $$ Also, $(T>t, t \leq a)=\emptyset$ if $t >a$ and $(T>t, t \leq a)=(T\leq t)^{c}\in \mathcal F_t$ if $t \leq a$.