If $X$ is a Lévy process, why is $t\mapsto\sum_{\substack{s\in[0,\:t]\\\Delta X_s(\omega)}}1_B(\Delta X_s(\omega))$ càdlàg?

Let $E$ be a normed $\mathbb R$-vector space, $(X_t)_{t\ge0}$ be an $E$-valued càdlàg Lévy process on a filtered probability space $(\Omega,\mathcal A,(\mathcal F_t)_{t\ge0},\operatorname P)$, $B\in\mathcal B(E)$ with $0\not\in\overline B$ and $$N_t(\omega):=\left|\left\{s\in(0,t]:\Delta X_s(\omega)\in B\right\}\right|=\sum_{\substack{s\in[0,\:t]\\\Delta X_s(\omega)}}1_B(\Delta X_s(\omega))$$ for $\omega\in\Omega$ and $t\ge0$.

How do we see that $t\mapsto N_t(\omega)$ is càdlàg?

In order for $N$ to be cadlag, it should be assumed that $X$ has finite jump activity in $B$, i.e. $\nu(B)<\infty$, where $\nu$ is the Lévy measure of $X$ (otherwise, it blows up to infinity immediately). And in such case, this is rather straightforward: with probability $1$, the number of jumps in $B$ is locally finite, and $N_t = \sum_{n\ge 1} \mathbf{1}_{[\tau_n,\infty)} (t)$, where $\tau_n$ is the time of $n$th jump in $B$; this is clearly cadlag.