If $f\colon\mathbb R\to\mathbb R$ is continuous, then the image of any compact set is compact. Especially, the image of $[0,1]$ is a compact set $C$. Then the image of $(0,1]$ is either $C$ or $C\setminus\{y\}$ for $x=f(0)$. The set $(0,1)$ is not of this form, hence cannot be the image of $(0,1]$.
For A, C, D one readily finds specific functions: $f(x)=0$, $f(x)=1-x$, $f(x)=|2x-1|$.
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 $S^1$.
Best Answer
$S^1$ is compact and $\mathbb R$ is not.