Given the following definition of a Lévy Process:
A Lévy Process on $\mathbb{R}^d$ is a $D([0,\infty);\mathbb{R}^d)$-valued random variable $X$ for which $\mathbb{P}\left(X(0)=0\right)=1$ and $$\mathbb{P}\left(X(t_k)-X(t_{k-1})\in B_k:1\leq k\leq n\right)=\prod_{k=1}^{n}\mathbb{P}\left(X(t_k-t_{k-1})\in B_k\right)$$ for any finite sequence $0=t_0<t_1<t_2<\ldots<t_n<\infty$ whenever $n\in\mathbb{N}$ and $B_1,\ldots,B_n\in\mathcal{B}(\mathbb{R}^N)$.
How do I deduce, directly, that the path of any Lévy process is stochastically continuous, i.e, $\lim_{s\to t}\mathbb{P}(|X(s)-X(t)|>\epsilon)=0$ ?
Any help is appreciated.
By the homogeneity of the increments, we have
$$\mathbb{P}(|X(s)-X(t)|>\varepsilon) = \mathbb{P}(|X(t-s)|>\varepsilon).$$
This means that
$$\lim_{s \to t} \mathbb{P}(|X(s)-X(t)|>\varepsilon) = 0$$
is equivalent to showing that $X(t-s)$ converges in probability to $0$ as $s \to t$. On the other hand, we know that $X$ has cadlag sample paths; this implies in particular continuity at $t=0$, i.e.
$$\lim_{r \downarrow 0} X_r =X_0 =0.$$
As pointwise convergence implies convergence in probability, this finishes the proof.