Under which conditions Mean Square Continuity implies Sample Continuity for Gaussian Processes? First, let us give the setting.
Let $(\Omega, \Sigma, \mathbf{P})$ be a probability space, let $T$ be some interval of time, and let $X: T \times \Omega \rightarrow S$ be a stochastic process.
By Mean Square continuity I mean:

Given a time $t \in T, X$ is said to be continuous in mean-square at $t$ if $\mathrm{E}\left[\left|X_t\right|^2\right]<+\infty$ and
$$
\lim _{s \rightarrow t} \mathbf{E}\left[\left|X_s-X_t\right|^2\right]=0
$$

One can check here also the definitions of continuity with probability 1 and sample continuity.
Q: I would like to know under which conditions for Gaussian Processes, continuity in mean square implies sample continuity.
I know that there exists the Kolmogorov Continuity Criterion that gives a sort of an answer. However, it requires an upper bound for the second moment to be bounded for all $s,t \in T$ where $T$ compact. A version for Gaussian Processes of this theorem is given here below:

Proposition. If ( $W_t: t \in T$ ) is a centered Gaussian process indexed by a compact set $T \subset \mathbb{R}^d$ with $\mathrm{E}\left|W_s-W_t\right|^2 \leq\|s-t\|^{2 \alpha}$, for all $s, t \in T$ and some $\alpha \in(0,1]$, then $W$ possesses a version with continuous sample paths such that $\left|W_s-W_t\right|=O\left(\|s-t\|^\alpha \log (1 / \| s-\right.$ $t \|)$ ), uniformly in $(s, t)$ with $\|s-t\| \rightarrow 0$, almost surely.

Now imagine that we have something of the form, $\forall \: s,t \in T$:
$$ \mathrm{E}\left|W_s-W_t\right|^2 \leq |s-t| + o(|s-t|)$$
Q: Can we still conclude that the paths are sample continuous?
Another way to rephrase the question in an interesting way is by using the following property of Gaussian Processes:

Theorem A Gaussian process $X$ on $T$ has continuous sample paths with probability one if, and only if, it is continuous at each fixed $t \in T$ with probability one; i.e.
$$P \left( \lim_{s \rightarrow t} X_s =X_t, \quad \forall t \in T \right) =1$$
if and only if
$$P\left(\lim_{s \rightarrow t} X_s=X_t\right)=1, \quad \forall t \in T$$

Q: Therefore it would be enough to know when, for Gaussian Processes, mean square continuity implies continuity with probability 1.
 A: I found an answer to this in "Weak Convergence and Empirical Processes" - van der Vaart, Wellner.
Sample path properties of a Gaussian process $W=\left(W_t: t \in T\right)$, such as boundedness, continuity or differentiability, are determined by the covariance kernel $K(s, t)=$ $\operatorname{cov}\left(W_s, W_t\right)$ of the process. Let $T$ be an index set and the square of the intrinsic semimetric be:
$$
\rho^2(s, t)=\operatorname{var}\left(W_s-W_t\right)=K(t, t)+K(s, s)-2 K(s, t) \text {. }
$$
Proposition (Continuity). If $\left(W_t: t \in T\right)$ is a separable, mean-zero Gaussian process with intrinsic metric $\rho$, then for any $\delta>0$,
$$
\mathrm{E}\left[\sup _{\rho(s, t) \leq \delta}\left|W_s-W_t\right|\right] \lesssim \int_0^\delta \sqrt{\log D(\epsilon, T, \rho)} d \epsilon .
$$
Where $D$ is the $\epsilon$-packing number. Furthermore, for $J(\delta)$ the integral on the right side,
$$
M^* := \mathrm{E} \sup _{\rho(s, t)<\delta} \frac{\left|W_s-W_t\right|}{J(\rho(s, t))}<\infty .
$$
Consequently, $\left|W_s-W_t\right|=O(J(\rho(s, t)))$, uniformly in $(s, t)$ with $\rho(s, t) \rightarrow 0$, almost surely.
$M^*$ is the modulus of continuity of subgaussian processes. An increasing function $\omega$ such that $\omega(0)=0$ is called a modulus of continuity for the random process $\left\{X_t\right\}_{t \in T}$ on the metric space $(T, d)$ if there is a random variable $K$ such that
$$
X_t-X_s \leq K \omega(d(t, s)) \quad \text { for all } t, s \in T \text {. }
$$
Evidently the function $\omega$ controls the "degree of smoothness" of $t \mapsto X_t$. To show that $\omega$ is a modulus of continuity, it clearly suffices to prove that
$$
K=\sup _{t, s \in T} \frac{X_t-X_s}{\omega(d(t, s))}<\infty \quad \text { a.s. }
$$
In the case prosed in my question one can verify that $D(\epsilon,T,\rho) \sim \frac{c}{\epsilon}$, for $c>0$ constant, so the process has indeed sample continuous paths.
