On the inductive construction of stopping times Suppose that $T_0 = \inf \{ t \ge 0 : X_t \in J\}$ is a stopping time.
Then I  would like to show that for $m=0,1,2,\dots$, 
$$T_{m+1} = \inf \{t > T_m: X_t \in J\}$$ is a stopping time. 
To do this, I think the following argument works. Assume $T_m$ is a stopping time. Then,
$$\{T_{m+1} \le t\}=\bigcup_{s\in \mathbb{Q}} \bigg(\{T_m < s<t\} \cap \{X_s \in J\} \bigg)$$from and since each set in the union of the RHS is in $\mathscr{F}_t$, so is the LHS.  QED.
I think there needs to be some additional assumptions on the paths of $X$(e.g. right or left continuous or both) and on the sets $J$ (e.g. closed or open).
Is this correct? Also,
how can we show this if $$T_{m+1} = \inf \{t \ge T_m: X_t \in J\}?$$
 A: I believe that you don't need further assumptions in your case, but generally you do need them, because your assumption that $T_0$ is a stopping time is already quite strong.
Assume $J$ is open. Assume $X$ is even continuous. If $T_{m+1}=t>T_{m}$, you have that for all $s \leq t$ (btw. I think you would need to adjust your union above a bit so that $s$ can be $t$) that $X_s \notin J$. Hence, LHS not equal RHS.
This is one example where one can see that additional assumptions are needed. But there is even more to it, mostly because the stopping time always depends strongly on the filtration and it is a lot easier if your filtration is right continuous. For this see "Brownian Motion and Stochastic Calculus, Karatzas and Shreve,1991" page 7 or chapter 1.2.
But in your case, I think my above argument would lead to a contradiction that $T_0$ is no stopping time, I would intuitively say, that, actually, everyting is correct what you did and no more assumptions are needed. If something is a stopping time, when it happens the first time, why not also if it happens the n-th time? If your filtration is not completely unnatural, like constant at some point in time...
For the last question, to me it looks like $T_0=T_m$ for all $m$ at the firts glance.
Hope this helped,
Cheers
