Adapting a solution to solve other similar problems (showing a set is open) I have looked at various proofs of subsets of complex numbers being open and the solutions are all different and look ad-hoc. I'm trying to find a general pattern that can at least solve most such problems. The following solution looks like one such template.
Problem: Show $S = \{z \in \mathbb C: 2 < |z - 2| < 4\}$ is open. 
Solution: Let $z \in S, \ r = \min\{|z - 2| - 2, 4 - |z - 2|\}$. 
Suppose $w \in N_r(z).$ Then $|w - z| < r$ meaning $|w - z| < 4 - |z - 2|$ and so $|w - 2| = |w - z + z - 2| \le |w - z| + |z - 2| < 4.$ 
Also, $|w - z| < |z - 2| - 2$. So, $2 < |z - 2| - |w - z| = |z - w + w - 2| - |w - z| \le  |z - w | + | w - 2| - |w - z| = |w - 2|$.
Thus $2 < |w - 2| < 4$ and so $w \in S.$
My question:
I came across this problem below
Proving $|z-1|<|z-i|$ is an open set
and was wondering if I can use the solution (of the fact $S$ is open) above as a template to solve this one. The magnitude of a complex number is real and so the problems look similar. I am not sure if what I did below works as I used two different variables to define $r$. If incorrect, can I modify my solution to make it correct, but also keep the general gist of the template?
Let $r = |z - i| - |w - 1|$. Suppose $w \in N_r(z)$. Then $|w - z| < |z - i| -|w - 1|$. So,  $|w - 1| \\ < |z - i| - |w - z| \\ = |z - w + w - i| - |w - z| \\ \le |z - w| + |w - i| - |w - z| \\ = |w - i|$
Thus $w \in \{z \in \mathbb C: |z - 1| < |z - i|\}$.
 A: One useful fact that solves many of these: if $f$ is a continuous function and $U$ is an open set, then $f^{-1}(U) = \{z: f(z) \in U\}$ is open. This works for functions from $\mathbb C$ to $\mathbb R$ as well as functions from $\mathbb C$ to $\mathbb C$ (in fact for functions from any topological space to any topological space).  The absolute value and real and imaginary parts functions are continuous, $z$ is continuous, constants are continuous, and 
sums and products of continuous functions are continuous.  So e.g. since $(0, \infty)$ is an open subset of $\mathbb R$ and $|z-i| - |z-1|$ is a continuous function from $\mathbb C$ to $\mathbb R$, $\{z:\; |z-1| < |z-i|\} = \{z:\; |z-i|-|z-1| \in (0,\infty)\}$ is open.
A: In the original problem, the given set $S = \{z \in \mathbb C: 2 < |z - 2| < 4\}$ is the region between two concentric circles, centered at $z=2$. The plan of the proof is to show that if we are given an $z\in S$ we can fid a radius $r$, depending on $z$ such that the open disk of radius $r$, centered at $z$ lies wholly within $S$.  What should $R$ be?  It depends on whether whether $z$ is closer to the inner circle or the outer circle, right?  That's where they got the minimum in their value for $r=r(z)$.  Again, I suggest you draw a picture.
Now we try to do the same thing for $$T=\{z\in \mathbb{C}: |z-1|< |z-i|\}.$$ $T$ is the set of points closer to $1$ than to $i$.  These are the points $z=x+iy$ such that $y<x$.  Suppose we choose such a point $z$.  As in the "template" we have to find $r=r(z)$ such that $N_r(z)\subset T.$  It's obvious, especially if you draw a picture, that the distance from $z$ to the line $x=y$ is the largest $r$ that will work.  Calculate this value of $r$ and then prove that it works. 
