= Solution
Independence gives $S_n\sim N(0,n)$. For every integer $m$,
$$
\mathbb E e^{2\pi i mU_n}
=\mathbb E e^{2\pi i mS_n}
=e^{-2\pi^2m^2n},
$$
since subtraction of an integer does not change the exponential. These are precisely the <Fourier coefficients> of the distribution on the circle.
One can also obtain a direct density proof on $[0,1]$, avoiding an endpoint issue in transferring circle convergence. Periodize the normal density:
$$
f_n(u)=\sum_{k\in\mathbb Z}\frac1{\sqrt{2\pi n}}
e^{-(u+k)^2/(2n)},\qquad0\le u<1.
$$
Its coefficients are the displayed Gaussian values, so its absolutely convergent <Fourier series> is
$$
f_n(u)=1+2\sum_{m\ge1}e^{-2\pi^2m^2n}\cos(2\pi mu).
$$
The sum of the absolute values of the nonconstant coefficients tends to zero, for example by domination by the summable sequence at $n=1$. Therefore $f_n\to1$ uniformly, and integrating any bounded continuous function on $[0,1]$ gives
$$
\boxed{U_n\ \xrightarrow{\mathrm d}\ \operatorname{Unif}[0,1].}
$$
In fact the convergence is in <total variation>.
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