As mentioned in the comments, we have to show that for each $t$, the random variable $X_t$ is Gaussian.
A first approach will be the following: we write for a fixed $t$,
$$X_t=\sum_{j=1}^n X_{ t\frac jn} - X_{ t\frac{j-1}n}=:\sum_{j=1}^n Y_{n,j} $$
and show that the sequence $\left(\sum\limits_{j=1}^n Y_{n,j}\right)_n$ satisfies Lindeberg's condition for the central limit theorem, i.e., we have to prove that
$$\tag{Lindeberg} \mbox{ for each positive } \varepsilon ,\quad n\mathbb E[X_{t /n}^2 \chi\{| X_{t /n}|\gt \varepsilon\} ]=0. $$
Claim. The sequence $(n\mathbb E[|X_{t/n}|^3 ] )_{n\geqslant n_0} $ is bounded, where $n_0$ is such that $t/n_0\lt\delta$.
To see this, we use Rosenthal's inequality with the independent and centered family of random variables $(X_{tj/n}-X_{t(j-1)/n} )_{j=1}^n $ and the exponent $3$. We get that
$$\sum_{j=1}^n\mathbb E|X_{tj/n}-X_{t(j-1)/n} |^3 \leqslant
K\mathbb E[|X_t|^3 ].$$
The term $\mathbb E[|X_t|^3 ]$ is finite even if $t$ is greater than $\delta$ (we write $X_t$ as a finite sum of increments $X_{t_i}-X_{t_{i-1} } $, and $t_i-t_{i-1}\leqslant\delta$). We thus obtain by stationarity of the increments that
$$n\mathbb E|X_{t/n}|^3\leqslant KC(t), $$
where $C(t)$ depends only on $t$. This proves the claim.
Lemma. Condition (Lindeberg) is satisfied if for each positive $\varepsilon$, $n\mathbb P(|X_{t/n}|\gt\varepsilon )\to 0$.
Indeed, we have for each $n$ and $R$,
$$n\mathbb E[X_{t /n}^2 \chi\{| X_{t /n}|\gt \varepsilon\} ]\leqslant \frac nR\mathbb E|X_{t/n} |^3 +R^2n\mathbb P(|X_{t/n} |\gt\varepsilon )$$
and by the claim,
$$n\mathbb E\left[X_{t /n}^2 \chi\{| X_{t /n}|\gt \varepsilon\}\right]\leqslant
\frac{ KC(t)}R +R^2n\mathbb P(|X_{t/n} |\gt\varepsilon ).$$
To conclude the proof, we notice that
$$1-\left(1-P(|X_{t/n}|>\varepsilon)\right)^n=P\left(|Y_{n,i}|>\varepsilon\text{ for some }1\le i\le n\right)\to 0$$
as $n\to\infty$ by the uniform continuity of $s\mapsto X_s$ on $[0,t]$. Thus
$n\log\left(1-P(|X_{t/n}|>\varepsilon)\right)\to 0$, but
$$n\log\left(1-P(|X_{t/n}|>\varepsilon)\right)\le -nP(|X_{t/n}|>\varepsilon)\le 0,$$
so we finally get $nP(|X_{t/n}|>\varepsilon)\to 0$, which completes the proof.