Complex Analysis - function on open unit disk with $\Re(f(z)) \geq 0$ 
Suppose $f:D \to \mathbb C$ is an analytic and non-constant function with $\Re(f(z)) \geq 0$ for all $z \in D$. Show that $\Re(f(z)) > 0$ for all $z \in D$, and that $\left|f(z)\right|\leq \frac{1+|z|}{1-|z|}$ for all $z \in D$ if given that $f(0)=1$. 

For the first part of the question, my attempt at a solution is brief: consider any $
z_0 \in D$ and any $0<r<\left||z_0|-1\right|$. By the Open Mapping Theorem, $f(B(z_0, r))$ is open in $\mathbb C$. But that means, assuming $\Re(f(z_0))=0$, that there must be some $\omega \in f(B(z_0,r))$ such that $\Re(\omega)<0$, since $f(B(z_0,r)) \cap \{z:\Re(z)<0\} \not = \emptyset$. Thus there is some $z_{\omega} \in D$ such that $\Re(f(z_{\omega}))<0$, a contradiction. Thus the first part in the statement is proved. 
For the second part of the question, however, I am not certain how to proceed. The setup looks very similar to the conditions for Schwartz' Lemma or the Schwartz-Pick Estimate, so I attempted to analyze $g(z)=f(z)-1$, unfortunately to no avail. Appreciate your input! 
 A: Consider the mapping $$T:w \mapsto \frac{w-1}{w+1} $$ which maps the half plane $\Re w>0 $ conformally onto the unit disk. The composition $$g=T \circ \frac{1}{f} $$ maps the unit disk into itself, and it fixes zero. Thus according to Schwarz' lemma, for all $z$ in the unit disk $$|g(z)| \leq |z| $$ which means  $$\left| \frac{1/f(z)-1}{1/f(z)+1} \right| \leq |z|. $$ Using the triangle inequality (as well as the reverse version) we also have $$\frac{1-|1/f(z)|}{1+|1/f(z)|} \leq |z| ,$$ and rearranging this yields the result.
A: Let $\phi(z)={1+z\over 1-z}$, which maps the unit disc $D$ conformally onto the right half plane with $0\mapsto 1$. Use Schwarz' lemma on the map $g(z)=(\phi^{-1}\circ f)(z)$ to get $|g(z)|\leq |z|$ for all $z\in D$. 
Now try to find out what $\phi$ does to a disc $D_r=\{z\in D : |z|<r\}$ for a given $0<r<1$. Recall that any Möbius transformation maps lines and circles to lines and circles. Since $\phi$ maps the real axis to the real axis, you get that $\phi(D_r)$ is a disc in the right half plane which is bisected by the line segment $\phi((-r,r))=\bigl(\phi(-r),\phi(r)\bigr)=\bigl({1-r\over 1+r},{1+r\over 1-r}\bigr)$. This disc is contained in the disc with center at $0$ and radius equal to $\phi(r)={1+r\over 1-r}$. 
It follows that if $|z|<r$ then $|f(z)|<{1+r\over 1-r}$. Letting $r\to|z|$, we get the desired result.
