Which conformal maps UHP$\to$UHP extend continuously to the closure? 
Does every conformal map of the upper half-plane $\{\text{Im }z >0\}$onto itself extend continuously to a map from its closure $\{\text{Im }z \geq 0\}$ to itself? If not, which ones do?

In this answer (written by a student, so there may be mistakes), some claims are made which I'm trying to prove. 
First: Any map $\varphi$ which has such an extension $\tilde{\varphi}$ is such that $\tilde{\varphi}(\mathbb{R})=\mathbb{R}$. Why? Certainly I see that $\tilde{\varphi}(\mathbb{R}) \supseteq \mathbb{R}$, but I don't see why they must be equal.
Second: In this case $\tilde{\varphi}(\mathbb{R})=\mathbb{R} \Longrightarrow \tilde{\varphi}$ is an FLT. In trying to show this, I've considered pre- and post-composing $\tilde{\varphi}$ with a map taking the UHP to the disk. Then the composition is analytic, takes the disk to the disk and preserves $S^1$. If I knew it were bijective, I could show that it is of the form $$e^{i\theta}\frac{z-a}{1-\bar{a}z},$$ but not knowing that it is bijective, I'm unsure of what to do.
 A: Boundary correspondence is tricky. I think it's better to show that the map is a  FLT, and only then  deal with the boundary. 
Let's begin with the unit disk $\mathbb D$. 
Claim 1. Every conformal map $f$ of $\mathbb D$ onto itself such that $f(0)=0$ is the rotation, $f(z)=e^{i\theta} z$. 
Proof: Apply the Schwarz lemma to $f$ and to $f^{-1}$; conclude that equality holds, use the equality statement in the lemma. 
Claim 2. Every conformal map $f$ of $\mathbb D$ onto itself is a  FLT.  
Proof: Apply Claim 1 to $g(z) = \dfrac{f(z)-f(0)}{1-\overline{f(0)}f(z)}$. 
Claim 3. Every conformal map $f$ of UHP onto itself is a  FLT. 
Proof: Apply Claim 2 to $g  = \varphi\circ f\circ \varphi^{-1}$ where $\varphi(z)=\frac{z-i}{z+i}$ maps the UHP onto $\mathbb D$. Use the fact that FLT form a group. 
Claim 3. Every conformal map $f$ of UHP onto itself is of the form $z\mapsto \dfrac{az+b}{cz+d}$ with real $a,b,c,d$. 
Proof. We already know that $f$ is a  FLT. Thus, it's continuous everywhere in $\mathbb C$ (except maybe one point), and the image of $\mathbb R$ is a circle or a line. As you know, it contains $\mathbb R$. Hence it is equal to $\mathbb R$. By the reflection principle, $\overline{f(\bar z)}= f(z)$, from where the conclusion about $a,b,c,d$ follows. 
But actually there is no need to mess with coefficients at this point. Just note that the symmetry $\overline{f(\bar z)}= f(z)$ implies that if $z_0$ is a pole of $f$, so is $\bar z_0$. But a FLT has at most one pole. So, either there is a pole on $\mathbb R$, or $f$ is  linear. 
