= Solution
The <L2-bounded martingale convergence theorem> states that a <martingale> with $\sup_n\mathbb E|X_n|^2<\infty$ converges both <almost surely> and with <convergence in L2> to a <square-integrable> <random variable>. Its hypotheses hold because $0<X_n<1$. Thus there is $X_\infty\in[0,1]$ <almost surely> with both kinds of convergence. Alternatively, the <martingale convergence theorem> gives the <almost sure convergence>, and the bound $|X_n-X_\infty|^2\leq1$ gives <convergence in L2> by the <dominated convergence theorem>.<Probability: Theory and Examples (fifth edition) by Rick Durrett>[§4.4, Theorem 4.4.6, printed p. 237: the Lp martingale convergence theorem]{ref}
The calculation of the <conditional expectation> of the square is especially informative. The <martingale> property makes the cross term vanish in the square expansion, so
$$
\mathbb E[X_{n+1}^2\mid\mathcal F_n]
=X_n^2+\tfrac14X_n(1-X_n).
$$
Subtracting this from $\mathbb E[X_{n+1}\mid\mathcal F_n]=X_n$ gives
$$
\mathbb E[X_{n+1}(1-X_{n+1})\mid\mathcal F_n]
=\tfrac34X_n(1-X_n).
$$
Taking <expected values> and iterating proves
$$
\mathbb E[X_n(1-X_n)]=x(1-x)(3/4)^n.
$$
By <bounded convergence theorem>, $\mathbb E[X_\infty(1-X_\infty)]=0$. The <random variable> inside this <expected value> is nonnegative, so it is zero <almost surely>; on $[0,1]$ this forces $X_\infty\in\{0,1\}$. Again by <bounded convergence theorem>, $\mathbb EX_\infty=\lim_n\mathbb EX_n=x$. Hence
$$
\boxed{X_\infty\sim\operatorname{Bernoulli}(x),\qquad
\mathbb P(X_\infty=1)=x,\quad\mathbb P(X_\infty=0)=1-x.}
$$
This is the <Bernoulli polarization martingale> with update size $1/2$. It also has an exact <mean squared error>: the terminal representation $X_n=\mathbb E[X_\infty\mid\mathcal F_n]$ and the <equality case for conditional second moments> give
$$
\mathbb E[(X_\infty-X_n)^2]
=\mathbb E[X_\infty^2]-\mathbb E[X_n^2]
=x(1-x)(3/4)^n.
$$
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