Assume we have a stochastic process $(X_t)_{t \geq 0}$ which is a local martingale in respect to some filtration $F=(F_t)_{t \geq 0}$. That means by definition that there exists an almost surely increasing sequence of stopping times $(T_k)_{k \in \mathbb{N}}$, that diverges almost surely and such that the stopped process $X^{T_k} = (X_t^{T_k})_{t \geq 0}$ with $X_t^{T_k} = X_{\min(t,T_k)}$ is a martingale for every $k \in \mathbb{N}$, e.g. for every $s \geq 0$ with $s \leq t$ it holds that $\mathbb{E}[X^{T_k}_t | F_s] = X^{T_k}_s$.
Now take another sequence of stopping times $(S_k)_{k \in \mathbb{N}}$ with $S_k \leq T_k$ and $S_k \rightarrow \infty$. Does it hold that the stopped process $X^{S_k}$ is a martingale too for all $k \in \mathbb{N}$? In other words, would the $S_k$ be a valid choice of stopping times in the definition for a local martingale?
I thought about applying the optional stopping theorem. I know that the stopped process for every $k$ is a martingale. Using this martingale in the optional stopping theorem and stopping again with a stopping time $S_n$ for some $n$ would yield that the stopped process $(X^{T_k})^{S_n}$ is again a martingale. But this seems to easy, since I don't need that the sequence of stopping times $S_k$ is smaller than $T_k$. Some help would be really appreciated or an counterexample, if the statement doesn't even hold.
This statement is correct and your proof works. We don't need $S_k \le T_k$ to say that $(X^{T_k})^S_k$ is a martingale. A slightly more general statement is that if $(S_k) \rightarrow \infty$ is a sequence of stopping times then $S_k \wedge T_k$ is also a localizing sequence. This is frequently used to make it so that the stopped process $X^{T_k}$ is a uniformly integrable martingale in the definition of a local martingale (by replacing $T_k$ with $T_k \wedge k$). In the case of continuous local martingales, we can also make $X^{T_k}$ a bounded martingale by defining $S_k := \inf\{t : |X_t| \ge k\}$ and replacing $T_k$ with $T_k \wedge S_k$. In the case that we also have $S_k \le T_k$ this would show $S_k = S_k \wedge T_k$ is also a localizing sequence.