= Solution
Construct a rate-$\lambda$ <Poisson process>, for $\lambda>0$, from independent waiting times $E_j$ with <exponential distribution> $\operatorname{Exp}(\lambda)$. Put $S_k=E_1+\cdots+E_k$, $S_0=0$, and $N_t=\max\{k:S_k\leq t\}$. The <strong law of large numbers> gives $S_k/k\to1/\lambda$, so there are only finitely many arrivals on each finite interval. Consequently the counting paths are <càdlàg>, and $N_0=0$ <almost surely>.
At any deterministic time $s$, the <memorylessness of the exponential distribution> says that the residual waiting time has again <exponential distribution> $\operatorname{Exp}(\lambda)$ and is independent of the observed history. Subsequent waiting times are fresh independent copies. Thus the <Poisson process> restarts independently at $s$, proving <independent increments> and <stationary increments>. Integrating the joint waiting-time densities over $0<s_1<\cdots<s_k\leq t<s_{k+1}$ gives
$$
\mathbb P(N_t=k)=e^{-\lambda t}\frac{(\lambda t)^k}{k!},\qquad k\geq0,
$$
so its time values have the expected <Poisson distribution>.
For $h\downarrow0$,
$$
\mathbb P(N_h\ne0)=1-e^{-\lambda h}\longrightarrow0.
$$
Together with <stationary increments>, this gives <stochastic continuity> at every time, from either side where applicable. All the <Lévy process> requirements hold, and
$$
\boxed{\mathbb E e^{i\theta N_t}=\exp\{\lambda t(e^{i\theta}-1)\},\qquad\psi(\theta)=\lambda(e^{i\theta}-1).}
$$
For $\lambda=0$, the identically zero process gives the degenerate case.
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