Let $(N_t)_{t\ge0}$ be an $\mathbb N_0$-valued Poisson process on a probability space $(\Omega,\mathcal A,\operatorname P)$. We know that if $N$ is almost surely right-continuous, then $N$ is almost surely nondecreasing. Thus, $$A:=\{\omega\in\Omega:N(\omega)\text{ is not nondecreasing}\}$$ is a $\operatorname P$-null set. Assume $N$ is càdlàg and note that $$B:=\{\omega\in\Omega:N(\omega)\text{ is not càdlàg}\}\supseteq A\tag1.$$ Since $\mathbb N_0$ is a closed subset of $\mathbb R$, $$\Delta N_t(\omega)=N_t(\omega)-N_{t-}(\omega)\in\mathbb N_0\;\;\;\text{for all }\omega\in\Omega\setminus B.\tag2$$
Let $k,T\in\mathbb N$. How can we show that $$\{\exists t\in[0,T]:\Delta N_t\ge2\}\subseteq\bigcup_{i=1}^{kT}\left\{N_{\frac ik}-N_{\frac{i-1}k}\ge2\right\}\tag3$$ on $\Omega\setminus B$?
In order to prove $(3)$, we may note that if $t\in[0,T]$, then there is an unique $i\in\{1,\ldots,kT\}$ with $$t\in\left[\frac{i-1}k,\frac ik\right)\tag4.$$ Now let $\omega\in\Omega\setminus B$ with $\Delta N_t(\omega)\ge2.$ Since $N(\Omega)$ is nondecreasing, I only obtain $$N_{\frac ik}(\omega)-N_{t-}(\omega)\ge N_t(\omega)-N_{t-}(\omega)=\Delta N_t(\omega)\ge2\tag5,$$ but I don't think that we necessarily need to have $N_{t-}(\omega)\le N_{\frac{i-1}k}(\omega)$.