Minimal value of probability according to the difference of a Levy-process Can we conclude for a Levy-Process, that for all $\epsilon>0$ it holds that 
$\min_{s\in [0,t]} \mathbb P\left(\left|X_t-X_s\right|\leq \epsilon\right)>0$?
Stochastic continuity doesn't seem to work, and the infinite divisibility property of $X_t$ may be useless since from the triangle inequality of $|\cdot|$ we get an estimate of the upper bound.
 A: By the stationarity of the increments, this is equivalent to $$\min_{s \in [0,t]} \mathbb{P}(|X_s| \leq \epsilon)>0$$ for all $t>0$.
Without any additional assumptions this does, in general, not hold true. Consider for example $X_t := t$ (i.e. a deterministic drift process), then $$\mathbb{P}(|X_t| \leq \epsilon) = 0$$ for all $t>\epsilon$.

Additional question: Does there exist for any $t>0$ some $\epsilon>0$ such that
$$\min_{s \in [0,t]} \mathbb{P}(|X_t-X_s| \leq \epsilon)>0 \quad \dots ?$$
Yes, that's true. As already mentioned above, we have by the stationarity of the increments
$$\min_{s \in [0,t]} \mathbb{P}(|X_t-X_s| \leq \epsilon) = \min_{s \in [0,t]} \mathbb{P}(|X_s| \leq \epsilon). \tag{1}$$
Since
$$\{|X_s| \leq \epsilon\} \supseteq \left\{\max_{s \in [0,t]} |X_s| \leq \epsilon \right\}$$
for all $s \in [0,t]$, we find
$$\min_{s \in [0,t]} \mathbb{P}(|X_s| \leq \epsilon) \geq \mathbb{P} \left( \max_{s \in [0,t]} |X_s| \leq \epsilon \right).$$
Since the Lévy process $(X_t)_{t \geq 0}$ has càdlàg sample paths, the random variable $Y:= \max_{s \in [0,t]} |X_s|$ is almost surely finite. By the montone convergence theorem (or the continuity of the measure $\mathbb{P}$),
$$\mathbb{P} \left( \max_{s \in [0,t]} |X_s| \leq \epsilon \right) \xrightarrow[]{\epsilon \to \infty} 1.$$
Consequently, we can choose $\epsilon>0$ (sufficiently large) such that
$$\min_{s \in [0,t]} \mathbb{P}(|X_t-X_s| \leq \epsilon) \geq \mathbb{P} \left( \max_{s \in [0,t]} |X_s| \leq \epsilon \right)>0.$$
