We have $$
M_t = \sup_{n \geq 1} \sup_{s \leq t} |X_s^n|
$$
We consider the processes $$
M^n_t =\sup_{m \leq n} \sup_{s \leq t} |X^m_s|
$$
For $n \geq 1$. We claim that $M^n_t$ are increasing adapted processes.
Indeed, for each $1 \leq m \leq n$, the processes $Y^m_t = \sup_{s \leq t} |X^m_s|$ are increasing (obviously) and adapted. Once you observe this, $M^n_t = \max_{m \leq n} Y^m_t$ is a finite maximum of adapted processes, hence increasing and adapted.
Now, for any (bounded) sequence of real numbers $a_n$, we have $\sup_{n \geq m} a_m \to \sup_{n \geq 1} a_n$. Therefore,$M_t^n \to M_t$ pointwise, which shows that $M_t$ is adapted as well.
Claim : Let $f:\mathbb R^+ \to \mathbb R$ be a right continuous function. Then $g(t) = \sup_{s \leq t} f(s)$ is a cadlag function.
Proof : Note that $g(t)$ is an increasing function, and therefore admits left and right limits at every point. It is therefore sufficient to prove that $g(t)$ is right-continuous.
Let $t_n \downarrow t$, we must prove that $g(t_n) \to g(t)$. Suppose not. Then, for some $\epsilon>0$, $g(t_n) > g(t)+\epsilon$ for all $n$. By the definition of the supremum, there exist $s_n \leq t_n$ so that $$f(s_n) > g(t)+\epsilon \geq f(t) +\epsilon$$ for all $n$.
We claim that $s_n > t$. Indeed, if $s_n \leq t$ then $f(s_n) \leq g(s_n) \leq g(t)$ which contradicts the choice $f(s_n)>g(t)+\epsilon$ made above. Consequently, by the squeeze theorem, $s_n \to t$ and therefore $f(s_n)\to f(t)$ by right-continuity, but we also know that $f(s_n)-f(t) > \epsilon$ for all $n$. This is a contradiction : consequently, $g$ is right continuous. $\blacksquare$
As $X^m_t$ is right-continuous, it follows from the above that $Y^m_t$ is cadlag. Finally, $M^n_t$, for any $n$, is a finite maximum over some of these cadlag functions, and is therefore cadlag itself (to see this, you can use induction, along with the formula $\max\{x,y\} = \frac{x+y+|x-y|}{2}$).
To show that $M_t$ is cadlag, we first note that with respect to $t$, it is an increasing process, because if $s \leq t$ then $M^n_s \leq M^n_t$ for each $n$ and you can take the limit. Therefore, the left and right limit functions $M_{s-}, M_{s+}$ are well-defined for sure, since every increasing function admits right and left limits at each point.
To show right-continuity, we must prove that $M_{s+} = M_s$ almost surely. Suppose not. In that case, on a set of non-zero probability, $M_{s+} \neq M_s$. In particular, for some $\epsilon>0$, the set $\{M_{s+} - M_s > \epsilon\}$ has non-zero probability. We note that $M_{s+} - M_s > \epsilon$ if and only if there is a sequence $s_n \downarrow s$ such that $M_{s_n} > M_s + \epsilon$ for every $n$.
However, by the definition of $M_{s_n}$ and the supremum, there exists $s'_n \leq s_n$ and a sequence of indices $m_n$ such that $$
|X^{m_n}_{s'_n}| > M_s+\frac{\epsilon}{2} \geq |X^{m_n}_{s}| + \frac{\epsilon}{2} \tag{1}
$$
for all $n$, which we may rearrange and write as $$
|X^{m_n}_{s'_n}| - |X^{m_n}_{s}| > \frac{\epsilon}{2} \tag{2}
$$
However, clearly $s'_n > s$ ,otherwise it would be absorbed under the supremum in the definition of $M_s$ and couldn't therefore admit a value bigger than $M_s$. It follows by the squeeze theorem that $s'_n \to s$. Also note that $$
|X^{m_n}_{s'_n} - X^{m_n}_s| \geq |X^{m_n}_{s'_n}| - |X^{m_n}_{s}| > \frac{\epsilon}{2} \tag{3(a)}
$$
and $$
|X^{m_n}_{s'_n} - X^{m_n}_s| \leq |X^{m_n}_{s'_n} - X_{s'_n}| + |X_{s'_n} - X_{s}|+ |X_s - X^{m_n}_s| \tag{3(b)}
$$
Claim : $m_n$ isn't a sequence of bounded numbers.
Proof : Suppose that $m_n$ is a bounded sequence of natural numbers. Then, there exists an $N$ such that $m_n \leq N$ for all $n$. Now, we know that $M^N_t$ is a cadlag process a.s. i.e. we know that for the sequence $s'_n$ above, we have $M^N_{s'_n} \to M^N_s$. However, note that $|X^{m_n}_{s'_n}| \leq M^N_{s'_n}$, so we know that $$\limsup_{n \to \infty} |X^{m_n}_{s'_n}| \leq \limsup_{n \to \infty} M^N_{s'_n} = M^N_{s}$$ However, from the first inequality in $(1)$, we get $$
\liminf_{n \to \infty} |X^{m_n}_{s'_n}| \geq M_{s}+\frac{\epsilon}{2} > M_s \geq M^N_{s} \geq \limsup_{n \to \infty} |X^{m_n}_{s'_n}|$$
This inequality above provides a contradiction, and completes the proof.$\blacksquare$
Therefore $m_n$ has a subsequence that converges to infinity. Let that convergent subsequence be $m_n$ itself, for the sake of easy notation.
In that case, $X^{m_n} \to X$ ucp. However, look now at the right hand side of $(3(b))$. The term $|X_{s'_n}-X_s|$ goes to $0$ as $n$ goes to infinity, as $X$ is cadlag. The terms $|X^{m_n}_{s'_n} - X_{s'_n}|$ and $|X^{m_n}_{s} - X_s|$ go to $0$ as $n \to \infty$ by the ucp convergence of $X^{m_n}$ to $X$. Finally, the entire right hand side of $(3(b))$ must go to $0$. However, this contradicts $(3(a))$, since $(3(a)),(3(b))$ combined tell you that the RHS of $(3(b))$ cannot go to $0$.
Therefore , if $(1),(2),(3(a))$ are to hold true on the set where $M_{s+} \neq M_s$, it must happen that either right-continuity or ucp convergence is violated at every element here. However, the intersection of both happens with probability $1$, therefore it is impossible that $M_{s+} \neq M_s$ be a set of non-zero probability.
Finally, $M_t$ must be a cadlag process.
Regarding the jumps of $M_t$, we see that $$
M_{t-} = \sup_{n \geq 1} \sup_{s < t} |X_s^n|
$$
Now, suppose that $M_t - M_{t-} =\epsilon> 0$. This means that the supremum in $M_t$ must occur at the time point $t$, so that means there is a sequence $m_n$ such that $|X^{m_n}_t| - M_{t-} > \epsilon-\frac 1n$ for each $n$. Note that $M_{t-} \geq |X^{m_n}_{t-}|$, so this implies via the triangle inequality that $|\Delta X^{m_n}_t| > \epsilon-\frac 1n$ for each $n$, or that $$\sup_{n} |\Delta X^n_t| \geq \epsilon = \Delta M_t = |\Delta M_t|$$ following the application of a limit and supremum on the LHS.
Best Answer
In the context of semimartingales, left limits come for free: a right-continuous finite variation process automatically has left limits, and ditto (a.s.) for a local martingales. There is probably some "historical accident" at play here too. Stochastic Calculus originally grew out of the theory of Markov processes, where cadlag was a default.