Continuity implying openess of a specific set I was looking at a proof on the Inverse Mapping Theorem using Rouche's Theorem and it had one step to which I can't seem to formulate a proof for. The argument in the step claims

Consider $f$ analitic. It is given that in the circle of radius $r$, the equation $f(z)=\alpha$, where $|\alpha| <\frac{r}{2}$, has only one solution. Let $U$ be the set (in $C_r$ such that $|f(z)|<\frac{r}{2}$. It follows that $U$ is open as $f$ is continious.

and is paraphrased from Serge Lange's Introduction to Complex Analysis at a graduate level

My thoughts:
I thought the proof of this would be something of the lines:
Let $z_0 \in U$. Let be $z_1$ in a neighbourhood of $z_0$ implying that $|f(z_0)-f(z_1)<\epsilon, \forall \epsilon >0$ due to continuity. Then, we have
$|f(z_0)|=|f(z_0)-f(z_1)+f(z_1)|\leq|f(z_0)-f(z_1)|+|f(z_1)|\leq \epsilon + \frac{r}{2}$
which does not work as we are not showing that this value is contained in the desired disc.
Question:
Does anybody have any ideas as how one could prove that $U$ is open?
EDIT: The whole proof can be found here  https://pasteboard.co/Kcvtsmf.png, https://pasteboard.co/KcvtOSE.png, https://pasteboard.co/Kcvuhg4V.png
 A: This is an application of a basic theorem about metric spaces and continuity, which can be found in any elementary book on metric spaces or on topology:

Theorem: For any function $f : X \to Y$ between two metric spaces, $f$ is continuous (using the $\epsilon,\delta$ definition of continuity) if and only if for each open subset $V \subset Y$, the set $f^{-1}(V) \equiv \{x \in X \mid f(x) \in V\}$ is an open subset of $X$.

In your case $X=Y=\mathbb C$ and $V = \{\alpha \in \mathbb C \mid |\alpha| < \frac{r}{2}\}$ which is obviously open, and $U = f^{-1}(V)$, so $U$ is open by applying the above theorem together with the fact that $f$ is continuous.
Something to keep in mind (which is also explained in elementary topology books): for any function $f : X \to Y$, the set valued inverse function is always defined, meaning that for any subset $V \subset Y$ its inverse image $f^{-1}(V)$ is a subset of $X$ defined by the formula given above.
A: $g(z)=|f(z)|, g: \Bbb C \to \Bbb R$, is a continuous function when $f$ is (and $f$ is even analytic). Then $$U = C_r \cap g^{-1}[(-\infty, \frac{r}{2})]$$ which is open as an intersection of open sets (assuming the disk $C_r$ is an open disk of radius $r$ so that $C_r = \{z\mid |z-z_0| < r\}$ for some $z_0\in \Bbb C$, say), using that continuity implies that the inverse image of an open set is open.
