I have a method but it assumes T and S are bounded stopping times. So let T and S be bounded stopping times. In Jacod and Protter, the theory is for Discrete time martingales and so T and S take on non negative integer values. (Note: All random variable equalities and inequalities are almost sure ones).
Let $M_t = E[Y|F_t]$
Then we need to show $ E[M_T|F_S] = M_{S\wedge T}$
Let $A= \{\omega : S(w) \leq T(\omega)\}$ $ T_1 = T\vee S$ $ T_2 = T \wedge S $ Note that $T_1$ and $T_2$ are stopping times. And $A \in F_S \cap F_T$. These are all exercises in Jacod and Protter.
$\therefore M_{S \wedge T} = M_S1_A + M_T1_{A^C}$
Now $E[M_T|F_S] = E[M_T 1_A + M_T1_{A^C}|F_S]$.
$E[M_T 1_A|F_S] = E[M_{T_1}1_A|F_S] = E[M_{T_1}|F_S]1_A$
But $T_1\geq S$. By Doob's Optional Sampling Theorem, and since $T, S, T_1$ and $T_2$ are bounded stopping times, $E[M_{T_1}|F_S]1_A = M_S1_A \quad (1)$
Similarly $E[M_T1_{A^C}|F_S]=E[M_{T_2}1_{A^C}|F_S]=E[M_{T_2}|F_S]1_{A^C}$
But $T_2 \leq S$. Therefore $F_{T_2} \subseteq F_S$.
$\Rightarrow E[M_T1_{A^C}|F_S] = M_{T_2}1_{A^C} = M_T1_{A^C} \quad (2)$
From (1) and (2)
$E[M_T|F_S] = M_S1_A + M_T1_{A^C} = M_{S \wedge T} $
You can interchange T and S and the proof goes through with the corresponding changes. As for the unbounded case, a limiting argument would be required. See if you can get it using the above, replacing T and S with $T\wedge N$ and $S \wedge N$, letting N go to $\infty$. Be careful though.
Also in "Stochastic Processes" by Richard Bass, the above is true even in the continuous time case.
Edit: Finally proved it for the unbounded case. But had to resort to Doob's Martingale convergence and Levy's zero one law, both available on Wiki.
Let T and S be two stopping times not necessarily bounded. Consider $ T_n = T\wedge n $ $S_n = S\wedge n$ Now $T_n$ and $S_n$ are bounded stopping times. Moreover $T_n \uparrow T$ and $S_n \uparrow S$. Consider the following: $E[M_{T_n}|F_{S_m}] = M_{T_n \wedge S_m} \quad (3)$ which is what we proved before. Now let us fix m. Let $T_n\wedge S_m = U_n$ and $T\wedge S_m = U$. Note $U_n \uparrow U$.
Claim: $M_{U_n}$ is a discrete martingale wrt $F_{U_n}$. And it satisfies the condition for Doob's martingale convergence.
Proof: Since $E[M_{U_n}] = E[M_0] = E[Y] < \infty$, and $E[M_{U_k}|F_{U_m}] = M_{U_m}$ if $m by Doobs OST and $\sup_{n \geq 1}E[M_{U_n}] = \sup_{n \geq 1}E[Y] <\infty$, the claim is proved.
Therefore $\lim_{n \rightarrow \infty} M_{U_n} = M_U$ Applying this to (3), we get $\lim_{n \rightarrow \infty} E[M_{T_n}|F_{S_m}] = \lim_{n \rightarrow \infty}M_{T_n \wedge S_m} = M_{T\wedge S_m}\quad (4)$
Now by the same logic, $M_{T_n} \rightarrow M_T$ and $\sup_{n \geq 1} E[M_{T_n}] = E[Y] < \infty $. By Conditional DCT, $\lim_{n \rightarrow \infty} E[M_{T_n}|F_{S_m}] = E[M_T|F_{S_m}] \quad (5) $ By (4) and (5) $E[M_T|F_{S_m}] = M_{T\wedge S_m}$ Now let $m \rightarrow \infty$. The RHS goes to $M_{T\wedge S}$ by Martingale convergence. Now Le'vy's conditional convergence (aka his zero-one law) states that if $X \in L^1$ and $F_m$ is a filtration then $E[X|F_m] \rightarrow E[X|F_\infty] = X$. Using this, we get $ E[M_T|F_{S_m}] \rightarrow E[M_T|F_{S_\infty}] = E[M_T|F_S]$
QED
Note: Le'vy's zero one law is a corollary of Doob's martingale convergence. So we as good as used that only and conditional DCT.