Here, we use the time-changed Brownian motion technique to show the normality of
\begin{align*}
Y_t = \int_0^t u\, dW_u,
\end{align*}
where $\{W_t, \, t \ge 0\}$ is a standard Brownian motion with respect to the filtration $\{\mathscr{F}_t,\, t \ge 0\}$. For $t\ge 0$, let $\mathscr{G}_t = \mathscr{F}_{\sqrt[3]{3t}}$. Consider the process $M=\{M_t, \, t\ge 0\}$, where
\begin{align*}
M_t = \int_0^{\sqrt[3]{3t}} u\, dW_u.
\end{align*}
Then, it is clear that $M$ is a continuous martingale with respect to the filtration $\{\mathscr{G}_t,\, t \ge 0\}$. Moreover, we have the quadratic variation $\langle M, M\rangle_t = t$. By Levy's martingale characterization of Brownian motion, $\{M_t, t \ge 0\}$ is a Brownian motion. That is, for $t> 0$, $M_t$ is normally distributed. Consequently,
\begin{align*}
Y_t &= \int_0^t u\, dW_u\\
&=M_{\frac{1}{3}t^3}
\end{align*}
is normally distributed, and $X_t = \frac{1}{t}Y_t$ is also normally distributed.
Comments
Note that, for $t>0$, $X_t \sim N\big(0, \frac{1}{3}t\big)$. Then, for any $\delta >0$,
\begin{align*}
\lim_{t \rightarrow 0} P(|X_t|>\delta) &=\lim_{t \rightarrow 0}2P(X_t > \delta)\\
&=\lim_{t \rightarrow 0}2P\left(\sqrt{\frac{3}{t}}X_t > \sqrt{\frac{3}{t}}\delta\right)\\
&=\lim_{t \rightarrow 0}\frac{2}{\sqrt{2\pi}}\int_{\sqrt{\frac{3}{t}}\delta}^{\infty}e^{-\frac{x^2}{2}}dx\\
&=0.
\end{align*}
That is, as $t$ approaches $0$, $X_t$ approaches $0$ in probability.