Is it true that a boundary of a simply connected and bounded set is connected in $\mathbb{C}$? I was thinking about Julia sets when I asked myself whether is it true that
Is it true that a boundary of a simply connected and bounded set is connected  $\mathbb{C}$?
Is it a well known fact?
I found a proof here: https://web.archive.org/web/20210506141049/https://esc.fnwi.uva.nl/thesis/centraal/files/f358061116.pdf but it's not convincing since it relies on facts that are not even proved nor mentioned...
Such that a path must cross the boundary in order to exit a set, the rather intuitive definition of "lying around" and the behavior of homotopy.
 A: Are you referring to Lemma 3.8? Then I agree that it seem odd how they construct the curve $\gamma$. 
I have a more direct approach showing sufficient path connectedness of the "neighbourhood" of the boundary. First we do the construct as in their proof of $\Omega = \{z\in\overline A: d(z, A^C)<\epsilon\}$ (this is an open set within $\overline A$ which if not obvious would be it soon), and if the boundary isn't connected this can be decomposed into two non-empty disjoint open sets $\Omega = \Omega_1\cup\Omega_2$. The path connectedness we need to show is that a point in $\Omega_1$ can be connected with a point in $\Omega_2$.
On order to construct such a path we can consider the function $f(z)$ being the distance to the complement, that is:
$$f(z) = \inf_{w\notin A} d(z,w)$$
this function is Lipschitz continuous with Lipschitz constant 1. By continuity it's now obvious that $\Omega$ is open, but moreover it means that $f$ is derivable almost everywhere which in turn means that the set of $d$ for which $f$ is derivable whenever $f(z)=d$ is dense. Furthermore $|f'|=1$ wherever it's derivable.
This means that we can create a set $\Gamma_d$ of level curves for each such $d$. Since $f$ is derivable with $f'\ne0$ on a such curve the curve cannot self intersect and therefore Jordan curve theorem applies. Since $A$ is simply connected $f(z)>d$ inside such a curve and only inside such a curve $f(z)>d$. Furthermore if $d<\epsilon$ these curves would have to run entirely within $\Omega$.
Now we select a point $w_1\in\Omega_1$ and $w_1\in\Omega_2$ with $f(\omega_1)=f(\omega_2)=d$ such that there is a set of level curves (this can be done since those $d$s form a dense set and by continuity $f$ will take all values between $\epsilon$ and $0$).
Now there's a level curve $\gamma_1\in\Gamma_d$ going through $w_1$ and $\gamma_2\in\Gamma_d$ going through $w_2$ and if it's the same we would be done. If it's not we simply form a path from $\gamma_1$ to $\gamma_2$ (which exists because since $A$ is simply connected it's path connected), and since $\gamma_2$ runs outside of $\gamma_1$ the path would be outside $\gamma_1$ and $\gamma_2$. 
The final step in constructing our final path is to replace any segment of it passing through the interior of another curve in $\Gamma_d$ is replaced by that curve. This way we have formed a path $\gamma$ such that $f(\gamma)\le d<\epsilon$, but since that we have $f(\gamma)\in\Omega$. 
So there's a path from $w_1$ to $w_2$ within $\Omega$.
