I try to plot the image of the domain $-\infty<x<\infty$ and $0<y<\pi$ under the function $w=\sqrt{z}$. I've got a hint that one boundary is a hyperbola.
I tried sth like that:
$w^2=z=x+iy \ \ \ \ \ w=u+iv \\ (u+iv)^2=u^2+2iuv-v^2 \\ u^2-v^2=x \ \ \ \ y=2vu \Rightarrow u={y \over 2v} \\ {y^2 \over 4v^2}-v^2=x \\ {y^2 \over4}-v^4=xv^2 \\ v^2={-x\pm \sqrt{x^2+y^2} \over 2 } \\ v=\pm {\sqrt{-x\pm \sqrt{x^2+y^2} \over 2}} \\ u=\pm {y \over 2\sqrt{-x\pm \sqrt{x^2+y^2} \over 2} }$
I'm not sure if it is a right approach to such kind of problem.
You went too far. Once you have $2uv=y$ then you know you have a hyperbola for any specific nonzero $y$ and of course the $u, v$ axes (to which the hyperbola are asymptotic) for $y=0$. So put in $y=0$ and $y=\pi$ with that information and get your boundaries.