Problem $13.7$ in Apostol's Mathematical Analysis. I am looking at problem $13.7$ in Apostol's Mathematical Analysis.

Let $D=\{|z|<1\}$ and let $f=u+iv$ be such that
$(\rm i)$ $u,v\in{\mathcal C}^1(D)$
$(\rm ii)$ $f$ is continuous on $\overline D$
$(\rm iii)$ $f(z)=z$ on $\partial D$.
$(\rm iv)$ $J_f(z)>0$ if $z\in D$.

Show that

$(1)$ For every open $\Omega\subseteq D$, $f(\Omega)$ is open in $f(D)$.
$(2)$ $f(D)$ is an open ball of radius $1$.
$(3)$ For each $w\in f(D)$, $D\cap f^{-1}(w)$ is finite.

I have only proven $(1)$:
PROOF1 Since $u,v$ are of class $\mathcal C^1$ over $D$, so is $f$, and since the Jacobian doesn't vanish, the inverse function theorem says that for each $z\in D$ there exists an open nbhd $N_z$ such that $f\mid_{N_z}$ is a diffeomorphism of class $C_1$, thus $f\mid_{N_z}$ is a homeomorphism, so it is an open map. Let $\Omega\subseteq D$ be open, and choose $z\in\Omega$. Then there exists $N_z$ as above. Let $\widetilde N_z=N_z\cap \Omega$. This is open and lies inside $N_z$ so its image under $f\mid_{N_z}$ is open by the above. Then $f(\Omega)=f(\Omega)\cap f(D)=\bigcup_{z\in\Omega}f\left(N_z\right)\cap f(D)$ is the intersection of the union of open sets with $f(D)$, so it is open in $f(D)$. $\blacktriangle$.
PROOF2 Suppose that $f(D)\not\subseteq D$. Then certainly $f(\bar D)\not\subseteq \bar D$. Thus, there exists $q\in\bar D$ such that $|f(q)|>1$. But $|f|$ is continuous on the compact $\bar D$; thus it attains its maximum on some $p\in\bar D$. But we cannot have $p\in\partial D$ since we would have $|f(p)|=1$, so it must be the case $p\in D$. Let $K=\overline B$ where $B=B(0,r)\; ;\; r=|f(p)|$. Then $f(\bar D)\subseteq \bar B$, and $f(p)\in\partial B$. Let $N_p$ be an open nbhd of $p$ in $D$. Then we see $f(N_p)$ is not open in $f(D)$, contrary to what was proven in $(1)$. It follows that $f(D)\subseteq D$.
PROOF3 Suppose for the sake of a contradiction that there exists $w\in f(D)=D$such that $D \cap f^{-1}(w)$ is infinite. Since this is a subset of the compact $\bar D$, it must have an accumulation point in $\alpha \in \bar D$. Suppose first that $\alpha\in D$. Then for each $\epsilon >0$ the ball $B(\alpha;\epsilon)$ contains infinitely many points of $\{z\in D:f(z)=w\}$, in contradiction with the fact $f$ is a local diffeomorphism in $\alpha$. Suppose $\alpha\in \partial D$. Then $f(\alpha)=\alpha$, and by continuity, $f(\alpha)=w$, which is absurd since $D\cap\partial D=\varnothing$. $\blacktriangle$.

Chapter $13$ is "Implicit functions and extremum problems". I am not sure what theorems to use to prove $(2)$, and I haven't thought about $(3)$, so I will worry about $(3)$ later. I want to prove $(2)$ now. Any hints?
ADD If $A$ is any set, $\partial A$ denotes its boundary and $\overline A$ its closure.
 A: Here's an approach to prove $f(D) = D$. First show that $f(D) \subset D$ as follows: The fact that $f(D)$ is open means that $|f|$ cannot attain a maximum on $D$ and must therefore attain a maximum on $\partial D$; use condition (iii) to conclude that $|f(z)|< 1$ for $z\in D$. Then show that $D\subset f(D)$, using the fact that $f(\overline{D})$ is a compact set that is at the same time equal to $f(D)\cup \partial D$. (Therefore, $\partial f(D)\subset \partial D$, meaning that $f(D)$ must contain every point of $D$.)

Here is a little more detail about the proof of $D\subset f(D)$, as requested in the comments. Note that $f(\overline{D}) = f(D)\cup \partial D$ is compact, hence closed, so it contains $\overline{f(D)} = f(D)\cup\partial f(D)$. As $f(D)$ is open, it is disjoint from its boundary, so $f(D)\cup \partial f(D)\subset f(D)\cup \partial D$ implies $\partial f(D)\subset \partial D$. Then $f(D)\cap D = \overline{f(D)}\cap D$, meaning that $f(D)\cap D$ is both closed and open relative to $D$; since $D$ is connected and $f(D)\cap D$ is nonempty, it must be the case that $f(D)\cap D = D$, or equivalently that $D\subset f(D)$.
A: Part 2
We know that $f(\bar D)$ is compact subset of the closed unit ball, so it is closed, and its complement in the closed unit ball is open. This follows because $f$ is an open mapping, so it obeys a maximum principle: its maximum modulus on a compact region cannot be taken in the interior. So we must have $|f|<1$ for all $z\in D$.
We can write $f(\bar D)^c$ in $\bar D$ as $\bar D\cap U$ for some open set $U$ in $\mathbb R^2$. Then $f(\bar D)^c\cap D= U\cap D$ is an open set of $D$ in the subspace topology. So the set of points in $D$ omitted by $f(D)$ in is open in the subspace topology on $D$, as the boundary points of $\bar D$ are all mapped to the boundary of $D$.
But this set is also closed. We know $f(D)$ is open by part 1, and its complement in the plane is closed. So $f(D)^c\cap D$ is a closed set of $D$ in the subspace topology.
We see the omitted set must be empty by connectedness, as it is both open and closed.
Part 3
If the inverse image of some point $w$ were infinite, the set of inverse images would have a limit point in the closed unit ball by compactness. If $p$ is the limit point, then by continuity $f(p)=w$. If $p$ is in the open unit ball, then no neighborhood of the limit point is mapped diffeomorphically onto an open set, contradicting the third hypothesis and the inverse function theorem.
If the limit point is on the boundary, we get a contradiction, as $w$ must lie in the open unit ball and $f(p)$ on the boundary.
