From $\lim_{s\to t}E e^{i \lambda\left(\zeta_{t}-\zeta_{s}\right)}=1$ prove $\zeta_t$ stochastically continuous at $t$. From
$$
\lim_{s\to t}E e^{i \lambda\left(\zeta_{t}-\zeta_{s}\right)}=1
$$
prove $\zeta_t$ stochastically continuous at $t$.

I don't know why it can be proven from such a condition.
If you need additional conditions, then $\zeta_t$ can assume to be a cadlag process with time-homogeneous and with independent increment.

What I can see from this condition is that $\zeta_s$ converges in distribution to $\zeta_s$ by Lévy's theorem.
But this may not imply converge in probability.
 A: Recall that if $X_n$ is a sequence of random variables (with values in $\mathbb{R}$), $a\in\mathbb{R}$, then $X_n\Longrightarrow a$ iff $X_n$ converges to $a$ in probability.
The statement in your posting means that $X_s=\zeta_t-\zeta_s$ converges to $0$ in distribution as $s\rightarrow t$. This in turn implies that $X_s$ converges to $0$ in probability as $s\rightarrow t$. All this means that for any sequence $s_n\xrightarrow{n\rightarrow\infty}t$, $\zeta_{t_n}$ converges to $\zeta_t$ in probability.
A: 
What I can see from this condition is that $\zeta_s$ converges in distribution to $\zeta_t$ by Lévy's theorem.

It proves more than that: it proves that $\zeta_s - \zeta_t$ converges in distribution to zero.  These are not the same thing!  For instance, suppose $X_n, X$ are iid standard normal; then $X_n \Rightarrow X$ in distribution (trivially, because they have the same distribution) but $X_n - X \Rightarrow N(0,2)$.
Now convergence in distribution to a constant implies convergence in probability, and $\zeta_s - \zeta_t \to 0$ i.p. does imply $\zeta_s \to \zeta_t$ i.p.
