To finally answer the question: This is a special case of the open mapping theorem for Riemannian surfaces. The theorem states that every holomorphic function $f:X\to Y$ between two Riemannian surfaces is an open mapping. If you apply it for $X=Y=\mathbb{C}\cup\{\infty\}$ you get that $f(G)$ is open in $Y$ for all open $G\subseteq X$. In particular the intersection $f(G)\cap\mathbb{C}$ is open because $\mathbb{C}$ is open in $Y$.
First note that openess is a local property: $f$ is open iff for all $x\in X$, $f$ maps (small enough) neighbourhoods of $x$ to neighbournoods of $f(x)$.
Now note that we can actually choose a small enough disc $U$ around $x$ such that $f(U)$ is contained in a disc $V$ around $f(x)$. Therefore we only have to show that holomorphic maps $\{z : |z|<r\} \to \{z : |z|<R\}$ that send $0$ to $0$ also map neighbourhoods of $0$ to neighbourhoods of $0$. Now this is either a consequence of the open mapping theorem for functions on the Gaussian plane if one has proven that previously or it is a relatively simple argument:
Write $f(z)=z^k \cdot g(z)$ for some $k\in\mathbb{N}_{\geq 1}$ and holomorphic $g$ with $g(0)\neq 0$.
First we show that we can reduce to the case $k=1$. By continuity $g(z)\neq 0$ for all $z$ in a sufficiently small neighbourhood of $0$. We can therefore write $g(z)=\exp(h(z))$ for some holomorphic function $h$. Now that means that $g$ has a holomorphic $k$th root: $g^{1/k}(z):=\exp(\frac{1}{k}h(z))$. Therefore we have $f(z)=(z\cdot g^{1/k}(z))^k$. Now the power map $w\mapsto w^k$ certainly maps open neighbourhoods of $0$ to open neighbourhoods of $0$. Therefore $f$ has this property if the inner function has it.
But in the case $k=1$ we have $f'(0) = 1\cdot g(0) + 0 \cdot g'(0)=g(0)\neq 0$ so that $f$ is locally a diffeomorphism around $0$ by virtue of the inverse function theorem. In particular it maps small enough open neighbourhoods of $0$ homeomorphically to open neighbourhoods of $0$. QED.
