Let $U:=\{z \in \mathbb{C} - \{0\} | -\pi < \arg z < \pi\}$.
For $z \in U$, the equation $w^2=z$ has two solutions $w=\pm(u + i v)$, where $u>0$.
Let $f : U \to \mathbb{C}$ be a function such that $f(z):=u+iv$, where $u>0$.
We define $\sqrt{z}:=f(z)$.
Prove that $\sqrt{z}$ is continuous.
Proof:
Let $z=x+iy\in U$ and $w^2 =(u+iy)^2= x+iy$ and $u>0$.
$u^2-v^2=x$ and $2uv=y$.
- If $y=0$, then $x>0$.
And $u=0$ or $v=0$.
And $u^2-v^2=x>0$.
So, $v=0$ and $u=\sqrt{x}$. - Suppose $y>0$.
$v=\frac{y}{2u}$.
$u^2-\frac{y^2}{4u^2}=x$.
$4u^4-4xu^2-y^2=0$.
$u^2=\frac{2x+\sqrt{4x^2+4y^2}}{4}=\frac{x+\sqrt{x^2+y^2}}{2}$.
$u=\sqrt{\frac{x+\sqrt{x^2+y^2}}{2}}$.
$v^2=u^2-x=\frac{-x+\sqrt{x^2+y^2}}{2}$.
$v=\sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}}$. - Suppose $y<0$.
$v=\frac{y}{2u}$.
$u^2-\frac{y^2}{4u^2}=x$.
$4u^4-4xu^2-y^2=0$.
$u^2=\frac{2x+\sqrt{4x^2+4y^2}}{4}=\frac{x+\sqrt{x^2+y^2}}{2}$.
$u=\sqrt{\frac{x+\sqrt{x^2+y^2}}{2}}$.
$v^2=u^2-x=\frac{-x+\sqrt{x^2+y^2}}{2}$.
$v=-\sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}}$.
So, for $z=x+iy \in U \cap \{x+iy \in \mathbb{C} | y\geq0\}$, $\sqrt{z} = \sqrt{\frac{x+\sqrt{x^2+y^2}}{2}} + \sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}} i$.
So, for $z=x+iy \in U \cap \{x+iy \in \mathbb{C} | y<0\}$, $\sqrt{z} = \sqrt{\frac{x+\sqrt{x^2+y^2}}{2}} - \sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}} i$.
Obviously $(x, y) \mapsto \sqrt{\frac{x+\sqrt{x^2+y^2}}{2}}$ is continuous on $\mathbb{R}^2$.
So, $(x, y) \mapsto \sqrt{\frac{x+\sqrt{x^2+y^2}}{2}}$ is continuous on $\{(x, y) | x+yi \in U\}$.
Obviously $(x, y) \mapsto \sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}}$ is continuous on $\mathbb{R}^2$.
So, $(x, y) \mapsto \sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}}$ is continuous on $\{(x, y) | x+yi \in U\cap\{x+iy \in \mathbb{C} | y>0\}\}$.
Obviously $(x, y) \mapsto -\sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}}$ is continuous on $\mathbb{R}^2$.
So, $(x, y) \mapsto -\sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}}$ is continuous on $\{(x, y) | x+yi \in U\cap\{x+iy \in \mathbb{C} | y<0\}\}$.
Let for $(x, y) \in U\cap\{x+iy \in \mathbb{C} | y\geq 0\}$ $g(x, y) := \sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}}$.
Let for $(x, y) \in U\cap\{x+iy \in \mathbb{C} | y< 0\}$ $g(x, y) := -\sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}}$.
Obviously $(x, y) \mapsto \sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}}$ and $(x, y) \mapsto -\sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}}$ are continuous on $\mathbb{R}^2$.
So, $(x, y) \mapsto \sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}}$ and $(x, y) \mapsto -\sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}}$ are continuous on $\{(x, y) | x+yi \in U\cap\{x+iy \in \mathbb{C} | y=0\}\}$ and $\sqrt{\frac{-a+\sqrt{a^2+b^2}}{2}} = -\sqrt{\frac{-a+\sqrt{a^2+b^2}}{2}} = 0$ for $(a, b) \in U\cap\{x+iy \in \mathbb{C} | y=0\}$.
So, $(x, y) \mapsto g(x,y)$ is continuous on $\{(x, y) | x+yi \in U\cap\{x+iy \in \mathbb{C} | y=0\}\}$.
It's easier to note that if $z = x + iy = re^{i\theta}$ then $\sqrt{z} = r^{1 \over 2}\cos{\theta \over 2} + i r^{1 \over 2}\sin{\theta \over 2}$, where $r^{1 \over 2} = \sqrt{x^2 + y^2}$ and $\theta$ (suitably defined) are both continuous functions of $x$ and $y$. Here you can use branches of $\theta = \arctan({y \over x}), \theta = \arccos({x \over \sqrt{x^2 + y^2}})$, or $\theta = \arcsin({y \over \sqrt{x^2 + y^2}})$ depending on the region.
Thus the real and imaginary parts of $\sqrt{z}$ are expressible as products of compositions of continuous functions and as a result $\sqrt{z}$ is continuous