 Solution

The printed update needs a one-step index correction to match its filtration: the draw for the transition from $n$ to $n+1$ must be the fresh variable $U_{n+1}$. Also the trivial initial sigma-algebra is $\{\varnothing ,\Omega \}$, rather than the printed $\{\varnothing ,\mathcal{F}\}$. With literal $U_n$, the next increment is already known at time $n$, so $\mathbb{E}[X_{n+1}\mid {\mathcal{F}}_n]=X_{n+1}$; the nontrivial update differs from $X_n$. Thus the stated martingale claim is false with that indexing. The following uses the standard Pólya urn described in words.

Put $m=B_0+R_0$, so the total count is $m+n$. Conditional on ${\mathcal{F}}_n$, the next blue indicator $I_{n+1}$ has expectation $X_n$. HenceAdaptedness and integrability follow from the recursive construction and $0\le X_n\le 1$. Thus $X_n$ is a bounded martingale. The martingale convergence theorem gives an almost-sure limit $X_{\infty }\in [0,1]$. Boundedness and the dominated convergence theorem then give $\mathbb{E}|X_n-X_{\infty }{|}^2\to 0$, establishing convergence in $L^2$.