Let $U$ be an open and connected subset of $\mathbb{C}$ and$f:\mathbb{D}\rightarrow U$ be a conformal equivalence, such that $f(0)=0$ and $g:\mathbb{D}\rightarrow U$ is an arbitrary holomorphic function with $g(0)=0$.
How can I show that $\forall_{0<r<1} r \; g(B_r(0))\subset f(B_r(0))$?
Note that if $\phi=f^{-1}\circ g$, then $\phi$ is a Schwarz map ($\phi$ holomorphic, maps the unit disc into itself and $\phi(0)=0$), so it satisfies Schwarz Lemma, while obviously by definition $g=f\circ \phi$ - or in other words $g$ is subordinated to $f$
But now if $0<r<1$ and $|z|<r$ is arbitrary in the $r$-disc, it follows that $\phi(z)=w$ satisfies $|w|<r$, while $g(z)=f(w)$, which is precisely what we needed to prove.