It is stated in a proof that there are only finitely many $s$ such that $|\Delta X_s|\geq\frac{1}{2}$ on each compact interval where $X$ is a càdlàg process.
I thought of a process with sample path recursvly defined as $X_i(\omega)=1$ for $i\in[0,1/2)$ and $X_i(\omega)=X_j(\omega)+1$ for $j\in[\sum_{i=1}^{n-1}\frac{1}{2^{i}},\sum_{i=1}^n\frac{1}{2^{i}})$ and $i\in[\sum_{i=1}^{n}\frac{1}{2^{i}},\sum_{i=1}^{n+1}\frac{1}{2^{i}})$ for $n\geq1$ and all $\omega\in\Omega$ (thus X is deterministic). Then we would have $\Delta X_s>\frac{1}{2}$ for infinitely many s on the compact interval $[0,1]$ and X should be càdlàg.
Let's take the compact interval to be $[0,1]$. Suppose there are infinitely many times $s\in[0,1]$ with $|\Delta X_s|\ge 1/2$. By compactness there is a convergent sequence of such times, and extracting a subsequence if necessary we can assume that the sequence is monotone, say monotone increasing. Thus we have $0\le s_1<s_2<\cdots\le 1$ with $s:=\lim_ns_n$, and $|\Delta X_{s_n}|\ge 1/2$ for each $n$. Because $X$ has left limits, for each $n\ge 2$ there is $t_n\in(s_{n-1},s_n)$ with $|X_{s_n}-X_{t_n}|\ge 1/4$. But this is absurd because $$ \lim_n X_{s_n}=\lim_n X_{t_n}=X_{s-}. $$