Sketch proof (relating the infinite product to the infinite sum via the inequality $e^x \geq 1 + x, \forall x$):

Since $x_t \leq -\log(1 - x_t), \forall t$ and $\sum_{t=1}^{\infty} x_t = \infty$, we have $\sum_{t=1}^{\infty} (-1)\log(1- x_t) = \infty$. Thus, $\prod_{t=1}^{\infty} (1-x_t)$ must either does not exist or converge to $0$ (if exists). Since $a_t := \prod_{\tau=1}^{t} (1-x_{\tau}), \forall t$ is a non-increasing sequence and bounded below (by $0$), thus $\lim_{t \rightarrow \infty} a_t$ exists which is then $0$.

Sketch proof (relating the infinite product to the infinite sum via the inequality $e^x \geq 1 + x, \forall x$):

Since $x_t \leq -\log(1 - x_t), \forall t$ and $\sum_{t=1}^{\infty} x_t = \infty$, we have $\sum_{t=1}^{\infty} (-1)\log(1- x_t) = \infty$. Thus, $\prod_{t=1}^{\infty} (1-x_t)$ must either does not exist or equals $0$ (if exists). Since $a_t := \prod_{\tau=1}^{t} (1-x_{\tau}), \forall t$ is a non-increasing sequence and bounded below (by $0$), thus $\lim_{t \rightarrow \infty} a_t$ exists which is then $0$.