Here is a proof coming from a french Math forum :
http://www.les-mathematiques.net/phorum/read.php?12,376956,377123#msg-377123
I translate the solution for the non french readers.
So here comes the solution (credit goes to egoroff) :
For $T(\omega)<\infty$, fix $\mathcal{H}$ as the collection of sets that do not separate $\omega$ and $\omega'$, i.e. sets $A$ s.t. either $\{\omega,\omega'\}\in A$ or $ \in A^c$. Then it is easy to see that $\mathcal{H}$ is a $\sigma$-field.
This was the first step. Next for every $(n+1)$-tuple $t_0<...<t_n\le T(\omega)$ and every Borel sets $A_{t_i}$, the set $(X_{t_i})_{i=0,...,n}\in \Pi_{i=0}^n A_{t_i}$ is in $\mathcal{H}$, by hypothesis over $\omega$ and $\omega'$, so $\mathcal{F}_t\subset \mathcal{H}$ for every $t\le T(\omega)$ as those set generate $\mathcal{F}_t$ .
Now $T(\omega)$ is known and finite we have :
-$S=T\wedge T(\omega)$ is a stopping time and moreover $S\in \mathcal{F}_{T(\omega)}\subset \mathcal{H}$ we have $S(\omega)=S(\omega')$, and so $T(\omega)\le T(\omega')$.
-On the other hand the event $\{T\le T(\omega)\}$ is in $\mathcal{F}_{T(\omega)}$, as $T$ is a stopping time so it is in $\mathcal{H}$, and $\omega\in \{T\le T(\omega)\}$ and so $\omega'$ too, and $T(\omega')\le T(\omega)$.
Finally we have shown that $T(\omega)=T(\omega')$ over $T(\omega)<\infty$ which was the claim to be proved.
Best regards
PS :
I also have a solution of mine based on a variant of Doob's lemma but as it is longer, more technical and far less elegant than this one, I do not post it here.
There are at least two reasons. I am certain of the second, only 60% of the first one, so please keep this in mind.
1. Look at equation (2.7); $F_t$ is a stopped version of the integral of the progressively measurable process.
It should follow that (at least restricted to locally bounded/integrable progressively measurable processes) that the integral itself is progressively measurable.
To the best of my understanding, an arbitrary stopped version of a measurable and adapted process does not necessarily have to be either measurable or adapted (or at least can fail one of the other two).
On the other hand, any progressively measurable function stopped at a stopping time is again progressively measurable. To the best of my understanding, the definition of progressively measurable also implies that progressively measurable processes are exactly the processes which are still adapted after stopping at deterministic times (and even progressively measurable still since deterministic times are stopping times).
2. I am 100% certain of this one. Look at the set $A$. As a result of its definition and the definition of progressive measurability, measurable and adapted is in general insufficient for $A$ to be measurable with respect to the product sigma-algebra; we need progressive measurability for this to be true. If $A$ is not measurable with respect to the product sigma-algebra, then Fubini is not applicable and we cannot say anything about the cross-sections of $A$, and therefore we cannot prove that there exists a sequence of simple predictable processes converging to $X_t$ in $L^2$.
Also, as a final note, even if adapted and measurable were sufficient for this proof (although I believe they are not), we still would only want to focus on progressively measurable processes anyway.
That is because the semimartingale stochastic integral only accepts locally bounded progressively measurable processes as integrands, so even if we could approximate in $L^2$ a wider class of functions by simple processes, we still would not be able to use that approximation in defining/generalizing stochastic integrals, so there would be no point in considering it.
Again though, I am fairly certain that progressive measurability is necessary for the proof to go through for either one of the two reasons mentioned above.
Best Answer
For every $n\in \Bbb N$ choose a subsequence $(m_k^n)_k$ in the following way.
For $n=1$ choose a subsequence $(m_k^1)_k$ such that $X_t^{(m_k^1)} (\omega) \to X_t (\omega)$ for almost every $(t,\omega ) \in [0,1] \times \Omega$.
Given the sequence $(m_k^n)_k$ we have that $(1)$ holds for $T=n+1$ and $(m_k^n)_k$. Thus, take $(m_k^{n+1})_k$ as a subsequence from $(m_k^n)_k$ such that $X_t^{(m_k^{n+1})} (\omega) \to X_t (\omega)$ for almost every $(t,\omega ) \in [0,n+1] \times \Omega$.
Now take as final sequence $m_k := m_k^k$, for which $$X_t^{(m_k)} (\omega) \to X_t (\omega)$$ holds for almost every $\cup_n [0,n]\times\Omega =[0,\infty) \times \Omega$.