this may be viewed as a duplicate of this post.
However i have put in much effort in the shared link and donated it with reputation, to check the proof considered there.
Here however i want to argue about another alernative proof, which may be usefull.
Lets say for a zero mean martingale $(X_{t})_{t\geq 0}$ and $Var[X_t]=1$ we have $$ \frac{X_t}{\sqrt{t}}\xrightarrow{d} \mathcal{N}(0,1) $$ (this is the case for mild assumptions of a process with stationary independent icrements). Lets say, if for a continuous increasing positive process $U_{t}$ and positive continuous increasing function $n(t)$ with $U_t,n(t)\rightarrow \infty$ as $t\rightarrow \infty$ it holds $$ U_{t}/n(t)\xrightarrow{P} Y\quad a.s. $$ for a strict positive and finite random variable.
I want to show that, $$ \frac{X_{U_t}}{\sqrt{U_t}}\xrightarrow{d}\mathcal{N}(0,1) \tag1 $$ holds.
Is it sufficient to show (1), that for every constant $0<\eta<\infty$ where now $$ U_{t}/n(t)\rightarrow \eta \quad a.s. $$ it holds, that $$ \frac{X_{U_t}}{\sqrt{U_t}}\xrightarrow{d} \mathcal{N}(0,1)\,.? $$ IF NOT here is the alternative: According to @Did notice we have that $$ g \left(\frac{X_{n(t)}}{\sqrt{n(t)}},\frac{U_t}{n(t)}\right)\xrightarrow{d} g (Z,\theta) $$ for any continuous function $g$ where $Z\sim\mathcal{N}(0,1)$ and $Z$ and $\theta$ are independent. Lets define $X_t \circ{} m:=X_{t\cdot m}$ we change the time. Then with $g(x,y)=\sqrt{y}^{-1/2}x\circ y$ we have $$ \frac{\sqrt{n(t)}}{U_t}\frac{X_{U_t}}{n(t)}\rightarrow \sqrt{\theta}^{-1/2} Z\circ \theta $$ It would be nice that the last expression is $\mathcal{N}(0,1)$ distributed. Since $\theta$ and $Z$ are independent we can think of a time scaled Brownian motion $Z=W_{1}$ where $W_{t}$ is a brownian motion at time t. Where for $\sqrt{\theta}^{-1}W_{\theta} \sim\mathcal{N}(0,\sigma^{2})$. Or could we use a better function $g()$?