Proving $|z-1|<|z-i|$ is an open set Consider this set:
$$|z-1|<|z-i|$$
Suppose $z=x+iy$, then:
$$\sqrt{(x-1)^2+y^2}<\sqrt{x^2+(y-1)^2}\implies$$
$$(x-1)^2+y^2<x^2+(y-1)^2\implies$$
$$x^2-2x+1+y^2<x^2+y^2-2y+1\implies$$
$$-2x<-2y\implies y<x$$
First of all, am I right?
Now, in order to prove that the set $O = \{(x,y); y<x\}$ is open, I need to pick a point $z\in O$, then construct an open ball of radius $r$ and prove it's entirely contained in $O$. I think that $r$ in this case must be the distance from this point $z$ to the line $y=x$. I have, then, to pick a point $w\in B(z, r)$ and prove that $w\in O$. How can I do that?

Note that the equation $|z - 1| < |z - i|$ says that that distance of $z$ from $1$ should be less than its distance from point $i$. If you plot the points $A = 1$ and $B = i$ on the argand diagram you see that $1$ lies on positive real axis and $i$ lies on upper imaginary axis each at a distance of $1$ from origin. First we need to find the points which lie at exactly equal distance from both points $A$ and $B$ and by basic geometry we know that these points form the perpendicular bisector of $AB$.
From the position of $AB$ is it clear that the perpendicular bisector passes through origin and is given by equation $y = x$. Now this line $y = x$ divides the whole complex plane into two parts: one part contains the point $A$ and other contains point $B$. We have to find points $z$ which are near to $A = 1$ compared to the point $B$. So the points $z$ lie in the part of the plane which contains $A$. Visually this portion lies below the line $y = x$ and given by the inequality $x > y$.
It is now almost obvious that the region below the line $y = x$ is an open set. But you require a formal / symbolic proof. You have also got the right idea but perhaps find it difficult to express it symbolically. Let $z = c$ be a specific point in the region below line $y = x$. Then we know that this point does not lie on the line $y = x$. Thus the distance $r$ of point $c$ from line $y = x$ is positive. Let's put $ c = a + ib$ where $b < a$. Then by coordinate geometry we know that $$r = \frac{a - b}{\sqrt{2}}$$ Now you have to consider the points $z$ which lie in a circle with center $c$ and radius $r$ i.e. points $z$ which satisfy $|z - c| < r$. We need to show that these points $z$ also lie below the line $y = x$.
Suppose $|z - c| = r'$ where $r' < r$. Then we can write $z - c = r'(\cos \theta + i\sin \theta)$ so that $$z = (a + r'\cos \theta) + i(b + r'\sin \theta) = x + iy\text{ (say)}$$ so that $$x = a + r'\cos \theta, y = b + r'\sin \theta$$ and then
\begin{align}
x - y &= a - b + r'(\cos\theta - \sin\theta)\notag\\
&= a - b + r'\sqrt{2}\cos(\theta + \pi/4)\notag\\
&\geq a - b - r'\sqrt{2}\notag\\
&= r\sqrt{2} - r'\sqrt{2}\notag\\
&= (r - r')\sqrt{2}\notag\\
&> 0 \text{ (because }r' < r)\notag
\end{align}
So the point $z = x + iy$ with $|z - c| = r' < r$ is such that $x > y$ and hence it lies below the line $y = x$. Since $c$ was an arbitrary point of the region $x > y$ it follows that every point in the region has a small ball type neighborhood all of whose points lie in the region $x > y$. It follows that the region given by $x > y$ is an open set.
Note that the above symbolic proof is totally unnecessary unless demanded by some insistent teacher/professor. One can use the the following obvious fact from geometry that if $\ell$ is a line and $P$ is a point not on $\ell$ and $Q$ is a point on $\ell$ such that $PQ$ is perpendicular to $\ell$ then all points inside the circle with center $P$ and radius $PQ$ lie on the same side of line $\ell$ as the point $P$. In other words the line $\ell$ divides the plane into two disjoint open regions.
