Analytic function from $\mathbb D$ to $\mathbb D$ that preserves the boundary is a rational function 
Assume that $f(z)$ is analytic on the open unit disk and continuous on the boundary $|z|=1$ with $|f(z)|=1$ if $|z|=1$. Show that $f$ is a rational function.


My attempt:
Since $f$ maps the boundary of the unit disk to the boundary of the unit disk, we can compose $f$ with some conformal map $\varphi$, such that $F\colon=\varphi^{-1}\circ f\circ\varphi$ maps the upper half-plane to the upper half-plane and preserves the real axis. For example, we may set $$ \varphi(z)=\frac{z-i}{z+i} .$$
Therefore, we can extend $F$ to an entire function with $F(z)=\overline{F(\bar z)}$ if $z$ lies in the lower half-plane. 
Now it suffices to show that $\infty$ is a pole then we invoke that every meromorphic function on $\overline{\mathbb{C}}$ is a rational function to conclude. But I am stuck... Any help?

Edit:
Can we claim that $$\lim_{z\to\infty}F(z)=\lim_{z\to x_0}z=x_0$$ where $x_0\in\mathbb R\cup\{\infty\}$ by tracing the images of the point which goes to infinity?
 A: Since $f$ does not vanish on $|z|=1$ it follows that it has at most finitely many zeros. Let $c_1,c_2,...,c_N$ be the zeros counted according to multiplicities. Consider $g(z)=\prod_{j \leq N} \frac {z-c_j} {1-\overset {-} {c_j}z }$. Observe that $|g(z)|=1$ when $|z|=1$. Verify that $h(z)=\frac {f{(z)}} {g(z)}$ is analytic on the open unit disk (ignoring removable singularities) and continuous on the closed disk with no zeros.   Apply MMP to $h$ and $\frac 1 h$ to conclude that $|h|$ is a constant. This implies that $h$ is a constant and completes the proof. 
A: Yes, your approach works. Just define  $F=\varphi^{-1}\circ f\circ\varphi$ on the extended upper half-plane 
$$\overline H = \{ z: \operatorname{Im} z \ge 0 \} \cup \{ \infty \} \subset \hat{\Bbb C} \,.
$$
That is possible because $f$ is continuous on $\overline{\Bbb D}$. Then extend $F$ to the lower half-plane via the Schwarz reflection principle, as you did. Now $F$ is meromorphic in $\hat{\Bbb C}$ and therefore a rational function.
Alternatively extend $f$ directly to a meromorphic function on $\Bbb C$ via reflection at the unit circle:
$$
 g(z) = \begin{cases}
f(z) & \text{ if } |z| \le 1 \\
\frac{1}{\overline{f(1/\bar z)}} &\text{ if } 1 < |z| < \infty  .
\end{cases}
$$
Then $\lim_{z \to \infty} g(z) = \frac{1}{\overline{f(0)}}$ so that $g$ has a removable singularity or a pole at $\infty$, and therefore is a rational function.
