A process $X$ is said locally bounded if there exists an increasing sequence $\{T_n\}$ of stopping times $T_n\rightarrow +\infty$ such that the stopped processes $$ X_t^{(T_n)} = \begin{cases} X_{t} & t\leq T_n\\ X_{T_n} & t\geq T_n \end{cases} $$ are all bounded, i.e. $\left|X_t^{(T_n)}\right|\leq K_n$, where $K_n$ is a constant (that may depend on $n$).
I often see that any adapted continuous process is proved to be locally bounded by choosing the sequence $$ T_n = \inf\{t\geq 0|\left|X_t\right|\geq n\}\quad(1) $$ This trivially gives $$ \left|X_t^{(T_n)}\right| = \begin{cases} \left|X_{t}\right|\leq n & t\leq T_n\\ n & t\geq T_n \end{cases}, $$ so that $\left|X_t^{(T_n)}\right|\leq n$ uniformly in $t$ and for all $n$. The question is: where do I need continuity? Sometimes I see that it is enough to have left-continuity. So I wonder whether we need continuity (or, perhaps, just left-continuity) just to have the time $T_n$ in equation $(1)$ well-defined.
UPDATE I found several results on this point. It is quite clear to me now that, in order to guarantee the hitting time $(1)$ to be a stopping time, we need to impose something to $X$. In particular, the set that is hit by $X$ is closed, so we can require the process $X$ to be continuous, see this LINK. I guess there are more general results, but they are difficult to be proved.