Why is there no continuous square root function on $\mathbb{C}$? I know that what taking square roots for reals, we can choose the standard square root in such a way that the square root function is continuous, with respect to the metric.
Why is that not the case over $\mathbb{C}$, with respect the the $\mathbb{R}^2$ metric? I suppose what I'm trying to ask is why is there not continuous function $f$ on $\mathbb{C}$ such that $f(z)^2=z$ for all $z$?
This is what I was reading, but didn't get:
Suppose there exists some $f$, and restrict attention to $S^1$. Given $t\in[0,2\pi)$, we can write
$$
f(\cos t+i\sin t)=\cos(\psi (t))+i\sin(\psi (t))
$$
for unique $\psi(t)\in\{t/2,t/2+\pi\}$. (I don't understand this assertion of why the displayed equality works, and why $\psi$ only takes those two possible values.) If $f$ is continuous, then $\psi:[0,2\pi)\to[0,2\pi)$ is continuous. Then $t\mapsto \psi(t)-t/2$ is continuous, and takes values in $\{0,\pi\}$ and is thus constant. This constant must equal $\psi(0)$, so $\psi(t)=\psi(0)+t/2$. Thus $\lim_{t\to 2\pi}\psi(t)=\psi(0)+\pi$.
Then 
$$
\lim_{t\to 2\pi} f(\cos t+i\sin t)=-f(1).
$$
(How is $-f(1)$ found on the RHS?) Since $f$ is continuous, $f(1)=-f(1)$, impossible since $f(1)\neq 0$.
I hope someone can clear up the two problems I have understanding the proof. Thanks.

Here is a proof for those who know a little complex function theory.  
Suppose $(f(z))^2=z$ for some continuous $f$.
By the implicit function theorem, $f(z)$ is complex differentiable (=holomorphic) for all $z\neq0$ in $\mathbb C$.
However since $f$ is continuous at $0$, it is also differentiable there thanks to Riemann's extension theorem.
Differentiating $z=f(z)^2$ at $z=0$ leads to  $1=2f(0)f'(0)=2\cdot0\cdot f'(0)=0 \;$. Contradiction.
