We proceed by weak induction on $N$. The case $N = 1$ is trivially true with $1 - x_1 \geq 1- x_1$. The case $N = 2$ is also true since $x_1 x_2 \geq 0$ and thus
\begin{equation}
(1 - x_1) (1 - x_2)
= 1 - x_2 - x_1 + x_1 x_2
\geq 1 - (x_1 + x_2)
\end{equation}
Suppose then that the claim holds for some $N \geq 2$. We have
\begin{equation}
\begin{split}
\prod_{n = 1}^{N + 1} (1 - x_n)
& = (1 - x_{N + 1}) \prod_{n = 1}^N (1 - x_n) \\
& \geq (1 - x_{N + 1}) \left( 1 - \sum_{n = 1}^N x_n \right) \\
& = 1 - \sum_{n = 1}^N x_n - x_{N + 1} \left( 1 - \sum_{n = 1}^N x_n \right) \\
& = 1 - \sum_{n = 1}^N x_n - x_{N + 1} + x_{N + 1} \sum_{n = 1}^N x_n \\
& \geq 1 - \sum_{n = 1}^{N + 1} x_n
\end{split}
\end{equation}
since $x_{N + 1} \sum_{n = 1}^N x_n \geq 0$. That is, the claim thus holds in the case $N +1$ and therefore the result is now a consequence of
R800: Proof by principle of weak mathematical induction. $\square$
