The Riemann mapping theorem says that there exsists a (bijective) conformal map $f$ between $\Omega =\Bbb C\backslash \{z\in\Bbb C: Im(z)=0, Re(z)\le0\}$ and the unit disc $D_1$.
$f$ is the composition of $\sqrt{z}$ and some Möbius transformation (I think $\frac{-iz-i}{iz-i}$) and $\sqrt z$ is not continuous.
But $f$ must be "biholomorphic" so holomorphic $\implies$ continuous
How can $f$ be continuous?
The set $\Omega$ is open and simply connected. It does not contain $0$. The principal value $\sqrt{\cdot}$ of the square root function, given by $$z\mapsto w:=\exp\bigl({1\over2}{\rm Log}(z)\bigr),$$ is analytic, hence continuous on $\Omega$. As is well known this function maps $\Omega$ bijectively onto the open half plane $H: \>{\rm Re}(w)>0$, and a suitable Moebius transformation will then map $H$ onto $D$.