Proceeding by induction, the base case of $N : = 1$ is vacuously true:
\begin{equation}
\sum_{n = 0}^N (x_{n + 1} - x_n) = (x_1 - x_0) = x_1 - x_0
\end{equation}
Assume then that the equality $\sum_{n = 0}^N (x_{n + 1} - x_n) = x_N - x_0$ holds for some $N \geq 1$. Then
\begin{equation}
\begin{split}
\sum_{n = 0}^{N + 1} (x_{n + 1} - x_n) & = (x_{N + 1} - x_N) + \sum_{n = 0}^N (x_{n + 1} - x_n) \\
& = (x_{N + 1} - x_N) + x_N - x_0 \\
& = x_{N + 1} - x_0
\end{split}
\end{equation}
The first claim therefore follows from
R800: Proof by principle of weak mathematical induction. For the second claim, suppose that $x$ converges to $0$ as $n$ increases without bound. Taking limits on both sides of $\sum_{n = 0}^N (x_{n + 1} - x_n) = x_N - x_0$ as $N \to \infty$, one has
\begin{equation}
\lim_{N \to \infty} \sum_{n = 0}^N (x_{n + 1} - x_n) = \lim_{N \to \infty} (x_N - x_0) = - x_0
\end{equation}
$\square$
