I have a question I am not sure whether my answer is correct or not:
I have a gaussian process $X_t$ (for $t\geq0$) and a random function $s(t):[0,\infty)\rightarrow[0,\infty)$.
Does $X_{s(t)}$ (for $t\geq0$) also a Gaussian process?
My answer is that it is because for every $\omega\in\Omega$, $s(t)$ will be a deterministic function and in that case, the process is Gaussian, and therefore it will be Gaussian in any case.
The problem is that this answer feels a little suspicious.
Let $X_t$ be standard Brownian motion, $\tau = \inf\{t : X_t \ge 1\}$ the hitting time of 1, and $s(t) = t \wedge \tau$. Then $X_{s(t)} \le 1$ almost surely so it is certainly not Gaussian.
The flaw in your argument becomes clear if you try to write it more precisely. $X_t = X_t(\omega)$ and $s(t) = s(t,\omega)$ both depend on $\omega$. You can't "plug in" an $\omega$ to $s(t)$ without also plugging it into $X_t$. And in that case you have $X_{s(t,\omega)}(\omega)$ which is not a process at all, it's a number, and it makes no sense to speak of it being Gaussian.
Let $\{X_t\}$ be a Brownian motion and let $\tau$ be a positive random variable independent of $\{X_t\}$ with $\mathsf{E}\sqrt{\tau}=\infty$. Then $$ \mathsf{E}|X_{\tau t}|=\mathsf{E}[\mathsf{E}|X_{\tau t}|\mid \tau]=\mathsf{E}\sqrt{\tau t}\mathsf{E}|X_1|=\infty. $$ Thus, $\{X_{\tau t}\}$ is not a Gaussian process.