 Solution

Because each increment is $\pm 1$, the event at time $k$ defining $T$ is exactly $\{X_{k-1}=X_k=1\}$. Therefore, for $n\ge 2$,and for $n<2$ this event is empty. Thus $T$ is a stopping time for the natural filtration.

On the other hand, $\{U\le 0\}=\{T=2\}=\{X_1=X_2=1\}$. This event has probability $1/4$, whereas the initial sigma-algebra ${\mathcal{F}}_0$ is trivial, even after completion up to null sets. It therefore cannot belong to ${\mathcal{F}}_0$. Hence $U=T-2$ is not a stopping time. Subtracting a deterministic delay from a stopping time can require information from the future.