Let $j \in J$. We have the decomposition
\begin{equation}
\{ \min(T_0, T_1) \leq j \} = \{ T_0 \leq j \lor T_1 \leq j \} = \{ T_0 \leq j \} \cup \{ T_1 \leq j \}
\end{equation}
Since $T_0$ and $T_1$ are stopping times, the sets $\{ T_0 \leq j \}$ and $\{ T_1 \leq j \}$ are each contained in $\mathcal{G}_j$. By definition, a sigma-algebra is closed under countable unions, which guarantees that the binary union $\{ \min(T_0, T_1) \leq j \} = \{ T_0 \leq j \} \cup \{ T_1 \leq j \}$ is in $\mathcal{G}_j$. This establishes the first claim.

As for the second claim, if $j \in J$, then we have the decomposition
\begin{equation}
\{ \max(T_0, T_1) \leq j \} = \{ T_0 \leq j \land T_1 \leq j \} = \{ T_0 \leq j \} \cap \{ T_1 \leq j \}
\end{equation}
Again, since $T_0$ and $T_1$ are stopping times, then the sets $\{ T_0 \leq j \}$ and $\{ T_1 \leq j \}$ are each contained in $\mathcal{G}_j$. Result

R1030: Sigma-algebra is closed under countable intersections shows that a sigma-algebra is closed under countable intersections, which guarantees that the binary intersection $\{ \max(T_0, T_1) \leq j \} = \{ T_0 \leq j \} \cap \{ T_1 \leq j \}$ is in $\mathcal{G}_j$. This concludes the proof. $\square$