In his treatise "An Introduction to Lévy and Feller Processes" (arXiv link), Prof. Dr. René Schilling gives a short and seemingly straightforward proof for the claim that a continuous Lévy process is integrable (Lemma 8.2 on p. 50). More precisely, the claim goes as follows.
Lemma Let $(X_t)_{t\geq0}$ be a Lévy process with càdlàg paths such that $|\Delta X_t(\omega)|:=|X_t-\lim_{s\uparrow t} X_s|\leq c$ for all $t\geq 0$ and some constant $c>0$. Then $E(|X_t|^p)<\infty$ for all $p\geq 0$.
The proof begins as follows.
Proof Let $\mathcal{F}^X_t:=\sigma(X_s,s\leq t)$ and define the stopping times $$ \tau_0:=0,\hspace{1cm} \tau_n:=\inf\left\{t>\tau_{n-1} : |X_t-X_{\tau_{n-1}}|\geq c\right\}. $$ Since $X$ has càdlàg paths, $\tau_0<\tau_1<\tau_2<\cdots$. Moreover, by the strong Markov property, $\tau_n-\tau_{n-1}\sim\tau_1$ and $\tau_n-\tau_{n-1}\perp\mathcal{F}^X_{\tau_{n-1}}$, i.e. $(\tau_n-\tau_{n-1})_{n\in\mathbb{N}}$ is an iid sequence.
The strong Markov property referred to in the proof is the following theorem (Theorem 4.12 on p. 29):
Theorem Let $X$ be a Lévy process on $\mathbb{R}^d$ and set $Y:=(X_{t+\tau}-X_\tau)_{t\geq0}$ for some a.s. finite stopping time $\tau$. Then $Y$ is again a Lévy process satisfying
a) $Y\perp(X_r)_{r\leq\tau}$, i.e. $\mathcal{F}^Y_\infty\perp\mathcal{F}^X_{\tau^+}:=\left\{F\in\mathcal{F}^X_\infty : F\cap\{\tau<t\}\in\mathcal{F}^X_t, \forall t\geq 0\right\}$.
b) $Y\sim X$, i.e. $X$ and $Y$ have the same finite dimensional distributions.
So the strong Markov property can only be applied in the lemma if $\tau_n$ is an a.s. finite stopping time, but why should this be the case?
So suppose, to avoid this caveat, we adjust the underlying probability space and consider only those $\omega$'s for which $\bigwedge_{k=1}^\infty \tau_k(\omega)\neq\infty$. Why should the correspondingly adjusted process $X$ remain Lévy, as required in order to apply the strong Markov property?
If, for all $T$, $P(|X_T-X_0|\leq 2c)=1$, then there is nothing to prove, the process is bounded so it is integrable.
If for some $T$ the above probability $\rho$ is less than $1$, then \begin{align}P(\tau_1=\infty)\leq P(\tau_1\geq NT)&\leq P(|X_T-X_0|\leq2c,|X_{2T}-X_T|\leq 2c,\cdots,|X_{NT}-X_{(N-1)T}|\leq 2c) \\&=\prod_{i=1}^N P(|X_{iT}-X_{(i-1)T}|\leq 2c) \\&=\rho^N,\end{align} by the property of i.i.d. increments, and since $N$ is arbitrary, $P(\tau_1=\infty)=0$.