How can I show that $T(omega) = T(overline{omega})$ when $X_{t}(omega)=X_{t}(overline{omega})$ for all $t in [0,T(omega)]cap [0,infty)$

The $sigma$-fields $mathcal F^X_t$ ($t>0$) have a "saturation" property that is the key: If $omegain Ainmathcal F^X_t$ and if $overlineomega$ is such that $X_s(omega) = X_s(overlineomega)$ for all $0le sle t$ then also $overlineomegain A$.

Now suppose we have $omega$ and $overlineomega$ such that $X_s(omega)=X_s(overlineomega)$ for all finite $s$ in $[0,T(omega)]$. Consider first the case $T(omega)<+infty$ and apply the saturation property with $t=T(omega)$ and $A={T=t}$. The conclusion is that $overlineomegain A$ as well; that is, $T(overlineomega) = t$.

To deal with the possibility that $T(omega)$ might be $+infty$, apply the preceding to the stopping time $T_n:=Twedge n$ ($n=1,2,ldots$) to conclude that $T(overlineomega)wedge n=T_n(overlineomega) = T_n(omega) =n$ for each $n$. Consequently, $T(overlineomega)=+infty=T(omega)$.