Uncorrelated white noise need not be iid

Every finite tuple of $X$ values is the same coordinatewise measurable function of a finite tuple of iid $U$ values. By [F1]–[F2], shifting all indices leaves the latter tuple's law unchanged, so the finite-dimensional laws of $X$ are shift invariant. Thus $X$ is strictly stationary.

Independence and [F3] give $\mathbb{E}[X_t]=\mathbb{E}[U_{t-1}]\mathbb{E}[1+U_t]=0$ and $\mathbb{E}[X_t^2]=\mathbb{E}[U_{t-1}^2]\mathbb{E}[(1+U_t{)}^2]=2$. For lag one, $\mathbb{E}[X_t{X}_{t+1}]=\mathbb{E}[U_{t-1}(1+U_t)U_t(1+U_{t+1})]=0$ because $U_{t-1}$ is independent of the remaining factors and has mean zero. For lag $h\ge 2$, $X_t$ and $X_{t+h}$ use disjoint iid coordinates, so they are independent and centered and their covariance is zero. Negative lags have the same covariance by symmetry. Hence $X$ is weakly stationary and is white noise with variance $2$.

Let $A=\{X_t=0\}$ and $B=\{X_{t+1}=2\}$. Since $U_{t-1}\in \{-1,1\}$, $A=\{U_t=-1\}$ and $\mathbb{P}(A)=1\mathrm{/}2$. The event $B$ requires $U_t=U_{t+1}=1$, so $\mathbb{P}(B)=1\mathrm{/}4$ while $A\cap B=\mathrm{\varnothing }$. Therefore $\mathbb{P}(B\mid A)=0\mathrm{\ne }\mathbb{P}(B)$, proving that adjacent observations are dependent.