Although our definition requires all analytic functions to be single-valued, it is possible to consider such multiple-valued functions as $\sqrt{z}$, $\log z$, or $\arccos z$, provided that they are restricted to a definite region in which it is possible to select a single-valued and analytic branch of the function.
For instance, we may choose for $\Omega$ the complement of the negative real axis $z\le 0$; this set is indeed open and connected. In $\Omega$ one and only one of the values of $\sqrt{z}$ has a positive real part. With this choice $w=\sqrt{z}$ becomes a single-valued function in $\Omega$; let us prove that it is continuous.
(The set $\Omega$ is the open set on which $f$ is defined.)
I don't really understand what all this means. Why is the negative real axis described by $z\le 0$ (shouldn't it be $x\le 0$?) Isn't it always the case that one value of $\sqrt{z}$ has a positive real part, since the two values are negative of each other? And why does $w=\sqrt{z}$ become a single-valued function, when we restrict the domain but not the range?