Since $X_0 : = Z = : X_1$, then clearly $X_0$ and $X_1$ agree in distribution. Furthermore, result
R3923: Standard gaussian random real number is symmetric about zero implies that
\begin{equation}
X_2 : = - Z \overset{d}{=} Z = : X_1
\end{equation}
This establishes the first claim. Next
\begin{equation}
\begin{split}
\mathbb{P}(X_0 \leq a, X_1 \leq a)
& = \mathbb{P}(Z \leq a, Z \leq a) \\
& = \mathbb{P}(\{ Z \leq a \} \cap \{ Z \leq a \}) \\
& = \mathbb{P}(\{ Z \leq a \}) \\
& = \mathbb{P}(Z \leq a) \\
& = \mathbb{P}(Z \in (-\infty, a]) \\
& \neq \mathbb{P}(Z \in [-a, a]) \\
& = \mathbb{P}(- a \leq Z \leq a) \\
& = \mathbb{P}(Z \leq a, Z \geq -a) \\
& = \mathbb{P}(Z \leq a, - Z \leq a) \\
& = \mathbb{P}(X_1 \leq a, X_2 \leq a)
\end{split}
\end{equation}
$\square$