The solution of a Ornstein-Uhlenbeck stochastic process is the following: $$X_t= m +(x_0-m)e^{-\lambda t }+ \sigma \int_0^te^{-\lambda(t-s)} \, dBs $$
we know that $ \sigma \int_0^te^{-\lambda(t-s)}\, dBs$ is a gaussian process. I also know that $X_t$ is a gaussian process and i thought that this is an immediate consequence of the fact that $ \sigma \int_0^te^{-\lambda(t-s)}\, dBs$ is a gaussian process and that adding to this term $m +(x_0-m)e^{-\lambda t }$ does not cause any problem. But actually it does, but I'm not able to understand why. So, why is not immediately visible that $X_t$ is a gaussian process applying the definition ( vec$(X_{t_1},.. X_{t_n})$ is multivariate normal for every $t_n$)?
Depending on the precise context, we might be able to use the following theorem:
For the proof, let $0 \leq t_1 < t_2 < \cdots < t_n \leq T$ and $a_1, a_2, \ldots, a_n \in \mathbb{R}$. Then $$\sum_{k=1}^n a_k\left(f(t_k)+Y_{t_k}\right) = \sum_{k=1}^na_kf(t_k) + \sum_{k=1}^na_kY_{t_k} = \text{constant } + \text{ normal random variable }$$ which is normal.
In your case, it seems that $f(t) = m + (x_0 - m)e^{\lambda t}$ is nonrandom and $Y_t = \sigma \int_0^te^{-\lambda(t-s)}\, dB_s$ was already proven to be a Gaussian process, so the theorem is immediately applicable.
Whether this result is "immediately visible" could be debated, I suppose, but since it's just a "throw-the-definition-at-it" proof, I don't know what problem your professor might be referencing.
I want to make one final comment: I haven't studied SDEs in a Bayesian context. If your class is Bayesian, then it would have been prudent for you to say so, and $f : [0,T] \to \mathbb{R}$ would likely need to be replaced by the random $f : [0,T] \times \Omega \to \mathbb{R}$. In this broader context, I have no clue if your $X_t$ even is a Gaussian process (my gut would say it isn't), much less how to prove it.