= Solution
The conditions on the <cumulative distribution function> imply $0<X_n\leq2$ <almost surely>. With $\mathcal H_n=\sigma(X_1,\ldots,X_n)$, adjoin $Y_0=1$ and the trivial $\mathcal H_0$. The <random variables> $Y_n$ are nonnegative, <adapted>, and bounded by $2^n$ at each fixed time. <Independence> and $\mathbb EX_{n+1}=1$ give
$$
\mathbb E[Y_{n+1}\mid\mathcal H_n]=Y_n\mathbb EX_{n+1}=Y_n.
$$
Thus this is the <product martingale from independent mean-one factors>, with $\mathbb EY_n=1$. The <almost sure supermartingale convergence theorem> states that a nonnegative <supermartingale> converges <almost surely> to a finite <integrable random variable>. Applied to this nonnegative <martingale>, it shows $Y_n\to Y_\infty<\infty$ <almost surely>. Every convergent real <sequence> is bounded, so \b[$\sup_{n\geq1}Y_n<\infty$ almost surely].
The printed maximum can in fact be taken literally here. If $X_1=1$ <almost surely>, all products equal $1$. Otherwise the <strict Jensen equality criterion>, applied to the <strictly convex function> $t\mapsto-\sqrt t$ on $(0,\infty)$, gives $q=\mathbb E\sqrt{X_1}<1$. <Independence> yields $\mathbb E\sqrt{Y_n}=q^n$. By <Markov's inequality>, $\sum_n\mathbb P(Y_n>\varepsilon)\leq\varepsilon^{-1/2}\sum_nq^n<\infty$ for every $\varepsilon>0$. The <First Borel-Cantelli lemma> and the countable choices $\varepsilon=1/m$ give $Y_n\to0$ <almost surely>. Since $Y_1>0$, a <sequence> with this limit attains its positive supremum among finitely many initial terms. This is an instance of <fractional moment decay of independent mean-one products>.
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