Square root of a simply connected region of $\mathbb{C}$. Let $\{0\} \subset B \subset \mathbb{C}$ be a connected, simply connected open subset of $\mathbb{C}$ containing $0$. 
I am trying to prove that $\sqrt{B}:=\{z \in \mathbb{C} \ | \  z^2 \in B\}$ is connected and simply connected without using the Riemann mapping theorem, but my attemps always stumble  over the fact that I would like to use a property of $B$ which is more or less equivalent to an instance of the (Riemann mapping) theorem.
Can anybody find a more elementary solution?
 A: Assume you know that a loop $I\to \Bbb C$ which is surjective on $B_{\epsilon}(0)$ can be homotoped relative to $\Bbb C \setminus B_{\epsilon}(0)$ to a path which is not surjective on $B_{\epsilon}(0)$. Then this path is also homotopic to a path which image does not intersect with $B_{\varepsilon/2}(0)$ relative to $\Bbb C \setminus B_{\epsilon}(0)$.
So you can homotop any path $\gamma \colon I \to \sqrt{U}$ to a path $\tilde\gamma  \colon I \to \sqrt{U}\setminus B_{\sqrt{\varepsilon/2}}(0)$. But $\tilde\gamma^2$ is homotopic to a path on the circle with radius $\varepsilon/2$. (Just look at a null homotopy of this path in polar coordinates and take the radius $r$ to be $\max(r,\varepsilon/2)$) Then since you know that $(\cdot)^2 \colon \Bbb C^{\times} \to \Bbb C^{\times}$ is a covering, by unique path lifting we can lift this homotopy to $\sqrt{U}$ and then we can easily show that this path is nullhomotopic. 
How to prove the first assumption? Look at $\gamma^{-1}(B_{\sqrt{\varepsilon}}(0))$ and $\gamma^{-1}(\Bbb C \setminus B_{\epsilon/2}(0))$ which are both unions of open intervals in $I$. Since $I$ is compact you know that there is a finite number of "subpaths" of $\gamma$ in $B_{\epsilon}(0))$ which cover $B_{\epsilon/2}(0))$. Since there is only a finite number of them, you can homotope them to straight lines and a finite number of straight lines does not cover $B_{\epsilon/2}(0))$.
