One can reduce it to the case of $f(0)=0$ (i.e. $0$ is the fixed point) by making a suitable linear fractional transformation of the disk. Namely, if $z^*$ is the fixed point, apply the above argument to $L \circ f \circ L^{-1}$ where $L$ sends $z^* \to 0$ and is a LFT.
Yes, the statement is true, and your solution is the standard argument. I'm not aware of a more elementary proof.
Let's write down the proof a little more systematically.
Since $f$ is by assumption injective (and analytic) on the boundary of the unit disk, $f\circ \gamma$ - where $\gamma$ is a positively oriented parametrisation of the unit circle - is a simple closed (continuously differentiable) curve, hence, by the Jordan curve theorem, it divides the plane into two disjoint domains whose common boundary is the trace of $f\circ\gamma$. Let $A_1$ denote the bounded domain (the interior of $f\circ\gamma$), and $A_2$ the unbounded domain (the exterior). Then the winding number
$$n(f\circ\gamma,z)$$
is constant on both domains, and $n(f\circ\gamma,z) = 0$ on $A_2$, while $n(f\circ\gamma,z) = \pm 1$ on $A_1$. But
$$n(f\circ\gamma,z) = \frac{1}{2\pi i}\int_{f\circ\gamma} \frac{dw}{w-z} = \frac{1}{2\pi i}\int_\gamma \frac{f'(\zeta)}{f(\zeta)-z}\,d\zeta$$
is, by the argument principle, the number of times $f$ attains the value $z$ in the open unit disk, hence non-negative, and it follows that $n(f\circ\gamma,z) \equiv 1$ on $A_1$. From $n(f\circ\gamma,z) \equiv 0$ on $A_2$, it follows that $f(D) \subset \overline{A}_1$, and by the open mapping theorem, it follows that $f(D)\subset A_1 = (\overline{A}_1)^{\Large\circ}$. Thus $f$ maps $D$ biholomorphically onto $A_1$.
Best Answer
I assume your function is analytic in the open unit disc and continuous on the closed unit disk. It has a finite number of zeros $z_j, j=1 \ldots n$ (counted by mutliplicity). We can extend the definition to the complement of the disc in the Riemann sphere by $f(z) = 1/\overline{f(1/\overline{z})}$ ($\infty$ if $f(1/\overline{z}) = 0$). Thus extended, you have an analytic function from the Riemann sphere to itself, and this must be a rational function (with poles at $1/\overline{z_j}$). Now $f(z) \prod_{j=1}^n \frac{1 - \overline{z_j} z}{z_j - z}$ has no poles or zeros and must be constant. We conclude that $f$ is a finite Blaschke product.