Automating proofs of the fact that $S \subset \mathbb C$ is open when topological continuity is unavailable There's a good method for proving a subset of complex plane is open that uses continuity. Right now this method is not formally available to me. The best method is to draw a picture and determine the radius of the neighborhood of an arbitrary complex number inside the given set, then use the triangle inequality or anything else appropriate for the situation to finish the proof. It takes up too much time and is very error prone. It sucks to have to do this when there's a more powerful, surefire method. Given this, I decided to make up a library of all the main complex subsets that are open, together with their proofs. So far I have


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*$|z - c_1| < c_2, \ |z - c_1| > c_2$

*$c_1 < |z - c_2| < c_3$

*$\Im(z) > c, \ \Re(z) < c$

*$|z - c_1| < |z - c_2|$
Since this is a naive solution to a slightly annoying problem, I have a few questions:


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*How feasible is this? This is not a technical question. It's just to see if the idea makes any kind of sense

*Let $X$ be a set of all open $S \subset \mathbb C$ of a kind (in a certain sense). If a proof method $Y$ can confirm $x_1 \in X$ is open, can $Y$ necessarily do the same for all $x \in X$? For example, if $Y$ can prove $|z| > 1$ is open, can $Y$ also do the same for, say, $|\arg(z)| > \frac{\pi}{2}?$ 
In case the answers to this question are "yes", "depends", "doesn't exist", "this question is inappropriate for this tag"

*What other cases can I add to the list of complex sets above?
 A: Automatic proofs do not exist, in my experience.
To illustrate this nonexistence, let me consider the example $|z-1| < |z-i|$ in your comment. 
My first step in trying to prove that this set is open is to fool around with the inequality, try to understand it in an intuitive, geometric manner, perhaps even try to formulate a good guess as to the nature of the solution set of the inequality. Once I feel that I have a good grasp on that, given $z$ in the solution set I would try to use my intuition to guess at an appropriate radius $r > 0$, expressed as a function $r=r(z)$, such that the open ball $B(z,r)$ around $z$ of radius $r$ is contained in the solution set of that inequality.
Armed with an actual formula for $r$ as a function of $z$, now my formal proof training kicks in, and I try to prove that if $|z-1| < |z-i|$, and if $|w-z| < r(z)$ then $|w-1| < |w-i|$, perhaps using "triangle inequalities or anything else appropriate".
So, what about the fun part? The fooling around? I sketch the complex plane. I plot the point $1$ and the point $i$. I plot a bunch of other points $z$ and ask myself: "Is this one closer to $1$ than it is to $i$? How about that one? How about this other one? ..."
Hopefully at some point, having done a lot of examples (or only just a few examples, if my luck and my experience hold), a lightbulb goes off in my mind: 

If $L$ is the perpendicular bisector of the line segment $\overline{1,i}$  then for all $z$, the inequality $|z-1| < |z-i|$ holds if and only if $z$ and $1$ are contained in the same half plane determined by $L$.

From this, I can now guess at a good formula for the radius $r=r(z)$: simply take the perpendicular distance from $z$ to $L$. This is not so hard, because I also have a good guess about the nature of the line $L$: it is just the line $y=x$. Also, I can probably prove at this point that $z=x+iy$ is in the solution set if and only if $y<x$ (proving that as a lemma might be a good idea).
As said, armed with that information, I can now write down the formal statement that must be proved: if $z=x+iy$, if $y<x$, and if $|w - z| < r(z)$, then $|w-1| < |w-i|$. I'll leave that to you, but with the final emphasis: you do need to know how to do these kinds of proofs, but you also need to allow yourself the fun part or else you'll never even be able to correctly formulate the necessary guesses. You must guess an appropriate formula $r(z)$! Without it, there's no way to proceed. And the best advice I have about how to find that formula is to use intuition, pictures, linear algebra, geometry, anything you have at hand.
