Functional limit theorem under random change of time FINAL EDIT: There is one main question left: According to the answer, we have choosen $\theta=1$ , where we could choose $0<\theta<\infty$ as we like. His this sufficient, if we regarde the convergence to $\theta$ as a finite positive random variable?
This post seems long, but its almost everything proofed except the last step. The unknown part is marked especially.
Assumptions
Given a Levy-Process $U_{t}$ with  with $E(U_t)=0$ (then $U_t$ is a martingale). Let $U_t$ have finite variance and $Var(X_1)=\sigma^{2}$ and the limit theorem  holds:
\begin{align}
F_t:=\sqrt{t}\left(\frac{U_t}{t}-E(U_1) \right)=\frac{U_t}{\sqrt{t}}\xrightarrow{d}\mathcal{N}(0,\sigma^{2})\quad as \,\,t\rightarrow \infty.\tag1
\end{align}
Let $K_t$ a non-decreasing positive ($K_{t}>0$ a.s.) process with cadlag-paths with the property that $K_{t}\rightarrow \infty$ almost sureley, as $t\rightarrow \infty$.
I want to show that 
\begin{align}
F_{K_t}:=\frac{U_{K_t}}{\sqrt{K_{t}}} \xrightarrow{d}\mathcal{N}(0,\sigma^{2})\quad as \,\,t\rightarrow \infty. \tag2
\end{align}
For this one requires a positive non-random cadlag-function $a(t)$ with $a(t)\rightarrow \infty$ as $t\rightarrow \infty$ such that
\begin{align}
\frac{K_{t}}{a(t)}\rightarrow \theta\quad P\, a.s. \tag3
\end{align}
holds. Where $\theta$ is a positive finite random-variable.
Then the convergence in distribution of $F_{t}\xrightarrow{d} \mathcal{N}(0,\sigma^{2})$ implies the convergence in distribution of $F_{K_t}\xrightarrow{d} \mathcal{N}(0,\sigma^{2})$.
The suggestion how it has to be proofed is given in the post below:
Here he takes $\theta=1$. So that we have $K_{t}\in ((1-\epsilon)a(t),(1+\epsilon)a(t))$ a.s. as $t\rightarrow \infty$. 
Is this legit, concerning that we have generally $\theta$ a positive ($\theta>0$) finite random variable?
For small $m$ we have
$$
P(U_{K_t}<x\sqrt{K_t})\leq P\left(K_{t}\notin ((1-\epsilon) a(t),(1+\epsilon) a(t))\right)+P\left(U_{a_t}<x\sqrt{(1+\epsilon)a(t)}+m\cdot \sqrt{\epsilon a(t))}\right)+ P\left(\sup_{s\in ((1-\epsilon)a(t),(1+\epsilon)a(t))}|U_{s}-U_{a(t)}|>m\cdot \sqrt{\epsilon a(t))}\right)
$$
The first term converges to 0 due to (3). The second term converges to $\Phi(x+m)$ (Why?) by the central limit theorem (1).
Due to the martingale maximale inequality the third term is bounded by 
$$\frac{1}{(m\cdot \sqrt{\epsilon a(t))})^{2}}$$ and tends to zero as $a(t)\rightarrow \infty$.
Why should this proof (2)? So far we have only that the distribution $P(U_{K_{t}}/\sqrt{K_t}\leq x)$ is bounded by $\Phi(x+m)$ then. A lower bound converging to $\Phi(x)$ is necessary i guess?
Idea:
$$
P(U_{K_t}<x\sqrt{K_t})\geq P(U_{K_t}<x\sqrt{K_t},|U_{a(t)}-U_{K_t}|<m\sqrt{\epsilon a(t)},K_{t}\in ((1-\epsilon)a(t),(1+\epsilon)a(t))\\
\geq P(U_{a(t)}<x\sqrt{(1-\epsilon)a(t)}-m\sqrt{\epsilon a(t)})
\\ \rightarrow \Phi(x-m)
$$
And we have it sandwiched, converging to $\Phi(x)$ right?
Btw: How to come to this inequality is due to
$$
P(U_{k_t}<x \sqrt{K_t})\\
\leq P[U_{k_t}<x \sqrt{K_t},K_{t}\in((1-\epsilon) a(t),(1+\epsilon) a(t))]+P[U_{K_t}<x\sqrt{K_t},K_{t}\notin ((1-\epsilon) a(t),(1+\epsilon) a(t))] \\ \leq P[K_{t}\notin ((1-\epsilon) a(t),(1+\epsilon) a(t))]+ P[U_{k_t}<x \sqrt{K_t},K_{t}\in((1-\epsilon) a(t),(1+\epsilon) a(t))] \\
\leq P[K_{t}\notin ((1-\epsilon) a(t),(1+\epsilon) a(t))]  \\
+P(U_{K_{t}}<x\sqrt{(1+ \epsilon)a(t)},|U_{K_t}-U_{a(t)}|\leq m\sqrt{\epsilon a(t)},|U_{K_t}-U_{a(t)}|> m\sqrt{\epsilon a(t)}]
\\ \leq P[U_{a(t)}<x\sqrt{(1+\epsilon)a(t)}+m \sqrt{\epsilon a(t)}] + P\left(\sup_{s\in ((1-\epsilon)a(t),(1+\epsilon)a(t))}|U_{s}-U_{a(t)}|>m\cdot \sqrt{\epsilon a(t))}\right)+P\left(K_{t}\notin ((1-\epsilon) a(t),(1+\epsilon) a(t))\right)
$$
 A: For square integrable Levy processes this should be easy because you can control the difference between $U_{K_t}$ and $U_{a(t)} $ with the martingale maximal inequality. Subtract out the mean and so suppose one has a square integrable independent increments process.  Then e.g. $$P(U_{K_t} < x \sqrt{K_t }) \le I_t+J_t+L_t$$ with $$I_t=P\left(K_t \notin ((1-\epsilon)a(t) , (1+\epsilon a(t))\right)$$ $$J_t= P\left(U_{a(t)} < x\sqrt{(1 + \epsilon) a(t)} + m \sqrt{\epsilon a(t)} \right)$$ and $$L_t= P\left(\sup_{s \in ((1-\epsilon)a(t) , (1+\epsilon a(t))} |U_s - U_{a(t)}| > m\sqrt{\epsilon a(t)}  \right) $$  $m$ should be small.  Use convergence is probability on $I_t$, CLT on $J_t$, and maximal inequality on $L_t$ (which will tell you that the sup can't be worse than $\frac 1 {(m \sqrt{\epsilon a(t)})^2}$).  Don't take every term literally as I am tex-ing on the fly, but that is the idea.
The assumption that $\frac{K_t}{a(t)}$ converges to a constant shows that $I_t$ is small. $J_t$ converges to $\Phi(x+m)$  by the CLT applied to $U_t$ or $U_{a(t)}$ as you prefer, and $L_t$ is  used as mentioned above.  Btw, I call this Anscombe's theorem.
