Let $a_{i+1} = 1-x_{i+1} = x_1x_2 \cdots x_i$ for $i \in \mathbb{N}.$ Then, we have that $a_1 = \frac12$, and for $n \ge 1$ we have that $a_{n+1} = x_1x_2 \cdots x_{n+1} = x_1x_2 \cdots x_n (1-x_1x_2 \cdots x_n) = a_n(1-a_n).$ We wish to show that $a_{100} \in (0.009, 0.01).$ It's easy to see by induction that $a_i < \frac1i$. Indeed, $a_n < \frac1n \Rightarrow a_{n+1} = a_n (1-a_n) \le \frac1n (1 - \frac1n) = \frac{n-1}{n^2} < \frac{1}{n+1}$ for $n \ge 2,$ and it's clear that $a_1 < \frac11$, $a_2 < \frac12$. This establishes that $a_{100} < \frac{1}{100} = 0.01.$ Now, we will tackle the $0.009$ part of the problem. Let $y_n = \frac1n - a_n.$ Then, we have that $y_1 = \frac12, y_2 = \frac14, y_3 = \frac{7}{48}, \cdots$ is defined by $y_{n+1} = \frac{1}{n+1} - (\frac1n - y_n) (1 + y_n - \frac1n) = \frac{1}{n^2(n+1)} + \frac{n-2}{n} y_n + y_n^2.$ The following lemma is the key. Lemma. $y_t < \frac{1}{t \sqrt{t}} \forall t \in \mathbb{N}.$ Proof. We will prove it by induction on $t$. For $t \le 9$, it's easily verified by simply using the recursion and doing some calculations. Suppose that it holds for $t = n$, where $n \ge 9.$ When $t=n+1$, we have that $$y_{n+1} \le \frac{1}{n^2(n+1)} + \frac{n-2}{n} \cdot \frac{1}{n\sqrt{n}} + \frac{1}{n^3}.$$Letting $a = \sqrt{n}$, we have that: $$\frac{1}{(n+1)(a+\frac{1}{2a})} < \frac{1}{(n+1)\sqrt{n+1}},$$so it suffices to prove that $$\frac{1}{n^2(n+1)} + \frac{n-2}{n} \cdot \frac{1}{n\sqrt{n}} + \frac{1}{n^3} \le \frac{1}{(n+1)(a + \frac{1}{2a})}.$$Rewriting this in terms of $a$, expanding, and then simplifying, we get that this is equivalent to: $$a^5 - 4a^4 + 5a^3 - 5a^2 + 2a-1 \ge 0.$$This rewrites as $$a^3(1 + (a-2)^2) - 4a^2 + 2a - 1 \ge 0,$$so since $a^3(1+(a-2)^2) - 4a^2 \ge 2a^3 - 4a^2 \ge 0$ (as $a \ge 3$) and $2a -1 \ge 0$ (as $a \ge 3$), we're done. $\blacksquare$ By the lemma, we have that $a_{100} = 0.01 - y_{100} > 0.01 - \frac{1}{100 \sqrt{100}} = 0.009,$ and so we're done. $\square$