Prove that a sequence converges but not to $0$

Note that $\log(1-x) \ge -x -{1 \over 2} x^2$ for $x \in [0,1)$.

$\log x_{n+1} = \log x_n + \log (1-{1 \over 2^{n+1}}) \ge \log x_n - ({1 \over 2^{n+1}} + {1 \over 2} {1 \over 2^{2n+2}})$.

Since the rightmost terms are summable, it follows that $\log x_n \ge -B$ for some $B>0$ and hence $x_n \ge {1 \over e^B}$ for all $n$.

An approach that doesn't require taking a logarithm like in copper.hat's nice answer: Notice that

$$x_n - x_{n + 1} = \frac{x_n}{2^{n + 1}}$$

Given the fact that $x_n > 2^{-n}$ (as can be seen by noting that almost all terms in the product are larger than $1/2$), we can iterate this identity,

$$x_0 - x_1 \ge \frac{1}{2^1}$$ $$x_1 - x_2 \ge \frac{1}{2^3}$$ $$x_2 - x_3 \ge \frac{1}{2^6}$$ and in general*,

$$x_k - x_{k + 1} \ge \frac{1}{2^{(k + 1)(k + 2) / 2}}$$

Combining these, we find that

$$x_0 - x_k \ge \frac 1 2 + \frac 1 {2^3} + \frac 1 {2^6} + \cdots + \frac 1 {2^{k(k + 1) / 2}}$$

By comparison to a geometric series, this difference is always less than $1 - 1/2^2 = 3/4$; hence, $x_n > \frac 1 4$ for every $n$.

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*The exponents are the $(k + 1)$-th triangular numbers. Convince yourself why.

Let $P_n$ be the product of the first $n$ terms. We show by induction that $P_n\ge\frac{1}{4}+\frac{1}{2^{n+1}}$.

Suppose the result is true for $n=k$. We show it is true for $n=k+1$. We have $$P_{k+1}\ge \left(\frac{1}{4}+\frac{1}{2^{k+1}}\right)\left(1-\frac{1}{2^{k+1}}\right)=\frac{1}{4}+\frac{1}{2^{k+1}}-\frac{1}{2^{k+3}}-\frac{1}{2^{2k+2}}.$$ To finish, it is enough to show that $\frac{1}{2^{k+3}}+\frac{1}{2^{2k+2}}\le \frac{1}{2^{k+2}}$. This is obvious, since $2k+2\ge k+3$.