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?

If you draw the picture, you see that your set is just $\mathcal O =\left \{ (x,y):y<x \right \}$ so pick a point $(x_0,y_0)\in \mathcal O$ and observe that the ball centered at $(x_0,y_0)$ of radius $\frac{\vert x_0-y_0\vert }{2\sqrt{2}}$ lies entirely in $\mathcal O$. 
Or if you know that $f(z)=|z-1|-|z-i|$ is continuous, then the result is immediate, since $\mathcal O=f^{-1}(\left \{ y\in \mathbb R:y<0 \right \})$.
