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?

Your set  consists of  the points in the plane that are closer to $1$ than $i$.
Consider the sequence $x_n=1/n$ you can see for every $n \in N$  it belongs to your set.
In fact every  natural number is in your set .Now through that sequence you can approximate zero which makes zero a limit point of your set (also belongs to the closure of the set)  but $0$  does not belong in your set .
So by definition if it was closed it should contain all of its limit points.Now this proves that it is not closed. So it can be neither open. So no idea on what to do maybe you can figure it out.
Another way to go is Consider the boundary of your  plane $y=x$  so now consider a point in your set $(a,b)$ what is its distance from  the line $y=x$ ? Now take a ball around $(a,b)$ with radius the half distance of the point $(a,b)$ from your line .It should be contained completely in your set. 
HINT:TO find the distance read this https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_line 
