I'm trying to solve the following question, but I have no idea why the hint was given as it was:
My attempt: I'm not really able to make use of the hint so far, so I'm a bit lost:
The assumptions give that $e^{i t x_n} \rightarrow c(t) + i s(t) \equiv z(t)$ for some real functions $s(t), c(t)$ with $\cos(x_n t) \rightarrow c(t)$, $\sin(x_n t) \rightarrow s(t)$.
$z(t)$ cannot be zero for any $t$ since $||e^{itx_n}| - |z(t)|| \leq |e^{itx_n} - z(t)|$ and $|e^{itx_n}| = 1$ so that $|z(t)| = 1$
$$E(e^{i t x_n U}) = f(x_n) \equiv \begin{cases} \frac{e^{itx_n} - 1}{itx_n} & x_n \ne 0 \\ 1 & x_n = 0 \end{cases}$$
Dominated convergence gives us that $$E(c(tU) + i s(tU)) = \lim_{n \rightarrow \infty} f(x_n)$$
Unfortunately I am not sure where to go from here.
The hint is suggesting that the sequence of random variables $(x_nU)$ is converging in distribution to some limit $L$. If you manage to prove this, it remains to deduce that the sequence $(x_n)$ itself is converging to something. This would be trivial if $(x_n)$ were bounded, in which case you conclude that $\frac12x_n=E(x_nU)\to E(L)$. But boundedness of $(x_n)$ is not given.
Instead, try proving that the sequence $(x_n)$ is Cauchy, i.e. that $$\lim (x_{n_k}-x_{m_k}) = 0$$ for any increasing sequences $(n_k)$ and $(m_k)$. By hypothesis $$\lim_k\exp (itx_{n_k}U)=\lim_k\exp(itx_{m_k}U)$$ almost surely. Therefore the ratio $$\exp(itx_{n_k}U) / \exp(itx_{m_k}U) = \exp it(x_{n_k}-x_{m_k})U$$ tends to $1$ almost surely. Apply dominated convergence to conclude $$E[\exp it(x_{n_k}-x_{m_k})U]\to 1$$ for every $t$, which implies $(x_{n_k}-x_{m_k})U$ converges in distribution to the constant $0$. From here you can argue the sequence $(x_n)$ is Cauchy.