We may write $$
Y_t(x) = 1_{(t+\mathbb{Q})\cap [0,1]}(x)
$$ and it follows $Y_t$ is $F_t=\mathcal{B}(\mathbb{R})$ measurable. And also, we can see that $
Y_t(x)= 0
$ holds almost surely, thus giving
$$
E[Y_t|F_s]=E[0|F_s]=0=Y_s
$$ holds almost surely for all $s<t$. This establishes that $\{Y_t,F_t\}$ is a martingale. Finally, we can observe that for all $x\in [0,1]$, the path
$$
t\mapsto Y_t(x)=1_{x+\mathbb{Q}}(t)
$$ of $Y_t$ is everywhere discontinuous.
To clear up a misconception, Kolmgorov's continuity theorem does not guarantee that there exists a set $A$ with $\mathbb{P}(A) = 1$ on which $t \mapsto X_t$ is continuous. Rather, it guarantees that there exists a continuous process $Y$ such that, for all $t \ge 0$, $\mathbb{P}(Y_t = X_t) = 1$. The important distinction is that $\{\omega : Y_t(\omega) = X_t(\omega)\}$ may be different for each $t$.
From what I understand of your question, you've shown that there exists a version $Y$ of $X$ satisfying $$\mathbb{E}\left[\left( \sup_{s \ne t \in [0,1]} \frac{|Y_t-Y_s|}{|t-s|^{\delta}}\right)^{\alpha} \right] < \infty$$ for all $\delta \in [0,\frac{\beta}{\alpha}[$, and what remains is to show that this version is Holder-continuous almost surely.
As a reminder, that means we want to show that for almost every $\omega$, we have that there exists constants $p = p(\omega) > 0$ and $C = C(\omega) < \infty$ (possibly depending on $\omega$) such that $|Y_t(\omega)-Y_s(\omega)| \le C |t-s|^p$ for all $t,s \in [0,1]$.
Using your definition of $k_\delta(\omega) := \sup_{s \ne t \in [0,1]} \frac{|Y_t(\omega)-Y_s(\omega)|}{|t-s|^{\delta}}$, we have immediately that $|Y_t(\omega)-Y_s(\omega)| \le k_\delta(\omega) |t-s|^\delta$ for all $t,s \in [0,1]$ just from the definition of $k_\delta$. Since $\mathbb{P}(k_\delta<\infty) = 1$, we conclude that we do indeed have that $Y$ is Holder-continuous almost surely. Even better, we have that it is almost surely $\delta$-Holder continuous for all $\delta \in [0,\frac{\beta}{\alpha}[$.
As an example to show we only have continuity of a modification, consider $\Omega = [0,1]$ with $\mathbb{P}$ equal to the Lebesgue measure, and define $X_t(\omega) = t1_{t \ne \omega}$. Then, for all $t,s \in [0,1]$, we have $\mathbb{E}[|X_t-X_s|^\alpha] = |t-s|^{\alpha}$, so Kolmgorov's continuity theorem applies and we can find a Holder continuous modification (namely $Y_t(\omega) = t$). However, for every $\omega \in \Omega$, the path $t \mapsto X_t(\omega)$ has a discontinuity on $[0,1]$, so there is no chance of finding a set with probability $1$ on which $X$ is continuous (Holder or otherwise).
Best Answer
One crude approach is to note that, writing $\|\cdot\|$ for the Euclidean norm on $\mathbb{R}^d$, we have $\|X\| \le d^{1/2} \max_i |X^i|$, and therefore $$\|Y_t - Y_s\|^\alpha \le d^{\alpha/2} \max_i |Y_t^i - Y_s^i|^\alpha \le d^{\alpha/2} \sum_i |Y_t^i - Y_s^i|^\alpha$$ and thus $$\mathbb{E} \|Y_t-Y_s\|^\alpha \le d^{\alpha/2} \sum_i \mathbb{E}|Y_t^i - Y_s^i|^\alpha \le C d^{\alpha/2} \sum_i |t-s|^{1+\beta} \le C d^{1+\alpha/2} |t-s|^{1+\beta}$$ so that $Y$ is Kolmogorov-continuous with parameters $\alpha, \beta, Cd^{1+\alpha/2}$ (the latter is probably not the best constant).
The independence is not needed.