Consider $D = \{z\in\Bbb C : |z|<1,\;\Re(z)>0\}$. Take $a\in D$ and consider $f$ a holomorphic function in $D$ such that $f(a)=a$ and $f(D)\subset D$. How can we prove that $|f'(a)|\le 1$?
I've been thinking about using Schwarz's lemma with $g(z) = f\circ T(z)$, with a Möbius transformation $T$ such that $T(0)=a$. We would have that $g(\Bbb D)\subset D \subset \Bbb D$ and $g$ would be holomorphic. So $|g'(0)|\le 1$ by the Schwarz lemma, but $g'(0) = f'(a)T'(0)$.
How do I find $T$ such that $T(0)=a$ and $T'(0)=1$? If this is not good, how else should I approach this?
There is no Möbius transformation which maps the unit disk $\Bbb D$ onto the half-disk $D$ (because Möbius transformations map disks to disks or half-planes).
But you don't need a Möbius transformation for this purpose, only some conformal mapping $\phi: \Bbb D\to D$ with $\phi(0) = a$. And that exists according to the Riemann mapping theorem because $D$ is simply connected.
Then $g = \phi^{-1} \circ f \circ \phi$ maps the unit disk into itself with $g(0) = 0$, so that the Schwarz lemma can be applied to $g$. It follows that $|g'(0)| \le 1$, and that leads to the expected result.
In other words: The conclusion $|f'(a)| \le 1$ holds for any function $f$ mapping a simply-connected domain $D$ into itself and has the fixed-point $a \in D$.