= Solution
Use the exponent convention $\mathbb E e^{iuX_t}=e^{t\psi(u)}$. A real <Lévy process> starts at $0$, has <stationary increments> and <independent increments>, is <stochastically continuous>, and is taken in its <càdlàg> version. Its <characteristic exponent of a Lévy process> satisfies $\psi(0)=0$ and is continuous. This convention has the opposite sign from the alternative $e^{-t\Psi(u)}$, so $\psi=-\Psi$.
For real $u$, the <exponential martingale of a Lévy process> is integrable because $|M_t^u|=e^{-t\operatorname{Re}\psi(u)}$ is deterministic and finite. For $s\le t$, its <conditional expectation> with respect to the natural <filtration> is
$$
\mathbb E[M_t^u\mid\mathcal F_s]
=e^{iuX_s-t\psi(u)}\mathbb E e^{iu(X_t-X_s)}
=e^{iuX_s-t\psi(u)}e^{(t-s)\psi(u)}=M_s^u.
$$
The second equality uses both <independent increments> and <stationary increments>. Thus this is a complex <martingale>, meaning that its real and imaginary parts are <martingales>.
A <probability distribution> is an <infinitely divisible distribution> when, for every positive integer $n$, it is the distribution of a sum of $n$ <independent and identically distributed> <random variables>. Their common distribution is allowed to depend on $n$. For a <Lévy process>,
$$
X_1=\sum_{j=1}^n\bigl(X_{j/n}-X_{(j-1)/n}\bigr),
$$
and these summands are <independent> with the common law of $X_{1/n}$. This proves the required <infinite divisibility (probability)> directly.
The <Lévy–Khintchine theorem> states that a real <infinitely divisible distribution> has <characteristic function> $e^{\psi(u)}$, where, for a unique triplet with the following fixed truncation,
$$
\boxed{\psi(u)=ibu-\frac a2u^2+\int_{\mathbb R\setminus\{0\}}
\bigl(e^{iuy}-1-iuy\mathbf1_{\{|y|<1\}}\bigr)K(dy),}
$$
with $b\in\mathbb R$, $a\ge0$, $K$ a positive measure, $K(\{0\})=0$, and $\int(1\wedge y^2)K(dy)<\infty$. Conversely every such triplet gives an <infinitely divisible distribution> and a <Lévy process>, unique in law, whose time-$t$ <characteristic function> is $e^{t\psi(u)}$. The <Lévy measure> $K$ need not be finite near zero. <Taylor expansion> makes the compensated integrand $O(y^2)$ there, while the measure is finite away from zero, so the integral is well defined. The drift parameter depends on the truncation: using $|y|\le1$ instead can alter $b$ by the contribution of atoms at $\pm1$. Here $a$ denotes <Gaussian> <variance>, not drift.
We now prove that continuity forces $K=0$, using the suggested truncation rather than merely quoting a path classification. For $\varepsilon>0$, $\lambda_\varepsilon=K(|y|\ge\varepsilon)$ is finite. The truncated exponent is that of the <Compound Poisson process>
$$
Y_t^\varepsilon=\sum_{j=1}^{N_t}J_j-t\int_{\varepsilon\le|y|<1}yK(dy),
$$
where $N$ has rate $\lambda_\varepsilon$ and the marks have law $K(dy)\mathbf1_{\{|y|\ge\varepsilon\}}/\lambda_\varepsilon$ when the rate is positive. The remaining exponent $\psi-\psi_\varepsilon$ also has an admissible <Lévy–Khintchine formula> triplet. Hence construct independently a <Lévy process> $Z^\varepsilon$ with that exponent. The sum $Y^\varepsilon+Z^\varepsilon$ has the same <finite-dimensional distributions> as $X$, and hence the same law on <càdlàg> path space. Continuity is therefore an almost-sure property of the sum if it is one of $X$.
But <independent compound Poisson jumps cannot cancel>: conditional on the path of $Z^\varepsilon$, its jump times are <countable>; the <independent> Poisson jump times have continuous distributions and avoid them almost surely. At each jump of $Y^\varepsilon$, the sum has the same nonzero jump, of size at least $\varepsilon$. If $\lambda_\varepsilon>0$, there is positive probability of such a jump on any fixed positive-length time interval, contradicting almost-sure continuity. Thus $K(|y|\ge\varepsilon)=0$ for every $\varepsilon>0$, and taking $\varepsilon=1/n$ gives $K=0$.
The <characteristic function> is now $\exp[t(ibu-au^2/2)]$. If $a>0$, the process $B_t=(X_t-bt)/\sqrt a$ has continuous paths and <independent> centered <normal> increments of <variance> $t-s$, so it is standard <Brownian motion> on the original probability space. If $a=0$, the increments are deterministic, and continuity together with equality at all rational times gives $X_t=bt$ simultaneously for all $t$ almost surely. A <Brownian motion> on an enlarged space can be used in the zero-variance representation if necessary. Therefore
$$
\boxed{X_t=bt+\sqrt a\,B_t,\qquad a\ge0.}
$$
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