Problem Suppose $f,g:\mathbb{D} \rightarrow \Omega$ are holomorphic functions from the unit disk to an arbitrary open set. Further assume that $f$ is bijective and $f(0) = g(0)$. Show that for any $0<r<1$ we have $g(B_r(0)) \subseteq f(B_r(0))$.
Due to the condition where $f(0)=g(0)$, my intuition says to maybe consider the function $h(z) = f(z) - g(z)$ so that $h(0) = 0$ and apply a lemma like Schwarz'. However, we don't necessarily have any idea of boundness of $|h(z)|$ so that is probably incorrect. What might be some other ways to show that it is a subset? Maybe I can define a biholomorphic function $p(z)$ that maps $\Omega$ into the open disk and study $p \circ f$ and $p \circ g$? Riemann Mapping theorem seem to not help here since we don't have any further conditions on $\Omega$ and we need simply connected and not all of $\mathbb{C}$ to use it.
Thank you.
EDIT: With the help of @user8675309 I make a new attempt:
Since $f$ is a continuous bijection, with the help of the open mapping theorem, this further implies that $\mathbb{D}$ is homeomorphic to $\Omega$. Thus, $\mathbb{D}$ and $\Omega$ share simply connectedness. Also, $\Omega$ must not be all of $\mathbb{C}$, otherwise $f^{-1}$ will be an entire function that is bounded so Liouville states this is a constant function but $f^{-1}$ is not constant. Therefore, we satisfy the conditions of the Riemann Mapping Theorem to obtain a biholomorphic function $p: \Omega \rightarrow \mathbb{D}$ such that $p(f(0)) = 0 = p(g(0))$.
Of course, the statement is now proven iff $(p \circ g)(B_r(0)) \subseteq (p \circ f)(B_r(0))$. By Schwartz', $|p(g(z))| \leq |z|$ and $|p(f(z))| \leq |z|$.
Now, if I can show that $|p(f(z))| = |z|$ then I am done. To show this, I either have to show that there exists $z_0 \neq 0$ such that $|p(f(z_0))| = |z_0|$ (OR $|p'(f(0))f'(0)| = 1$ but we don't have an explicit formula for $p'$ and $f'$ to study this case). I want to maybe consider, for the sake of concrete computation, the closed ball with radius $1/2$ centered at the origin: $B:= \overline{B_{1/2}(0)}$ and maybe apply the maximum modulus principle. However, the modulus may not necessarily grow linearly? Although I am suspecting that $(p \circ f)(z) = e^{i\theta}z$.
Applying Schwarz Lemma to $f^{-1}\circ g$ which maps the unit disc to itself and $0 \to 0$ (the map is well defined since $f$ is onto $\Omega$), one gets$|f^{-1}\circ g(z)| \le |z|$ which means that if $|z| \le r$ and $g(z)=w$ one has $|f^{-1}(w)=u| \le r$ so $w=f(u), |u| \le r$ which is precisely the statement to be proved