Let $\{ \mathcal{F}_s \}_{s \in [0, \infty)}$ be the pullback filtration for $X$. Fix $t \in [0, \infty)$ we have
\begin{equation}
\{ \min(T_0, T_1) \}
= \{ T_0 \leq t \text{ or } T_1 \leq t \}
= \{ T_0 \leq t \} \cup \{ T_1 \leq t \}
\end{equation}
Since $T_0$ and $T_1$ are stopping times for $X$, then both $\{ T_0 \leq t \}, \{ T_1 \leq t \} \in \mathcal{F}_t$. By definition, a sigma-algebra is closed under countable unions, which guarantees that the union $\{ T_0 \leq t \} \cup \{ T_1 \leq t \}$ is also in $\mathcal{F}_j$. This establishes the first claim.

As for the second claim, we have the factorization
\begin{equation}
\{ \max(T_0, T_1) \leq t \}
= \{ T_0 \leq t \text{ and } T_1 \leq t \}
= \{ T_0 \leq t \} \cap \{ T_1 \leq t \}
\end{equation}
Since sigma-algebras are closed under countable intersection, we conclude that $\max(T_0, T_1)$ is a stopping time using analogous reasoning as before. This concludes the proof. $\square$