Existence of continuous angle function $\theta:S^1\to\mathbb{R}$

You have actually proved the hard direction. The other direction is easy: if $U$ is not all of $S^1$, it misses at least one point $z\in S^1$. Without loss of generality, this point is $1=e^0$. Then define $\theta(e^{it})=t$, where $0<t<2\pi$. This function is continuous at every point of $S^1$ except $1$.

Proof. Suppose that $z_n=e^{i t_n}$ converges to $z=e^{it}\in S^1 -1$, but that $t_n$ doesn't converge to $t\in (0,2\pi)$. Since $t_n\in (0,2\pi)$, $t_n$ must have a convergent subsequence to some $t^*\in [0,2\pi]$ different from $t$. Note $0<|t-t^*|<2\pi$. Since $e^{iz}$ is continuous, we must have that $e^{i(t-t^*)}=e^0=1$. But $z=2\pi$ is the least nonzero real for which $e^{iz}=0$, so we've arrived at a contradiction.

I will provide a proof of a related fact. Try to see if you can modify it to prove what you want: suppose $V$ is an open subset of $\Bbb C$ and $f$ is a branch of the logarithm in $V$. (By definition, this is a continuous function $f:U\to \Bbb C$ for which $e^{f(z)}=z$. One can show that this guarantees $f$ is (complex) differentiable.) Then $S^1\not\subseteq V$.

Proof. Suppose $S^1\subseteq V$. Since $e^{f(z)}=z$, $f'(z)=1/z$. But then $$0= \frac{1}{2\pi i}\int_{S^1} f'(z) dz = \frac{1}{2\pi i}\int_{S^1} \frac{dz}z=1$$

Note that if you had a continuous map $u:S^1\to\Bbb R$ for which $e^{i u(z)}=z$, we would be able to define $f:\Bbb C^\times \to\Bbb C$ by $f(z)=\log |z|+ u(z|z|^{-1})$. This is continuous, and would furnish a branch of the logarithm in all of $\Bbb C^\times$, which is an open set containing $S^1$.

Let $\theta (z)=\Im (\log z)$ where $\log z$ is the branch for which $0<\arg z<2\pi$. This should work unless I'm missing something.