Show that $\hat{E}\setminus E$ is homeomorphic to $S^1$ 
Set $\mathbb{D}=\left\{z\in\mathbb{C}: \lvert z\rvert <1\right\}$ and define $\mathcal{P}\colon\mathbb{D}\times\mathbb{D}\to\mathbb{R}$ by
  $$
\mathcal{P}(x,y):=\begin{cases}\frac{1-\lvert x\rvert^2}{\lvert x-y\rvert^2}, & x\neq y\\0, & x=y\end{cases}.
$$
  (The function $\mathcal{P}$ is known as the Poisson kernel.) Let
  $$
E:=\left\{0\right\}\cup\bigcup_{n\in\mathbb{N}}\left\{(1-2^{-n})e^{\pi ik/2^n}: k\in\left\{0,1,\ldots,2^{n+1}-1\right\}\right\},~~\mathcal{F}:=\left\{\mathcal{P}(x,\cdot): x\in E\right\}
$$
  Show that the boundary $\hat{E}\setminus E$ of $E$ with respect to $\mathcal{F}$ is homeomorphic to $\mathbb{S}^1:=\left\{z\in\mathbb{C}: \lvert z\rvert=1\right\}$. Here $\hat{E}$ denotes the compactification of $E$ with respect to $\mathcal{F}$.

We had the following way to construct the compactification $\hat{E}$ of $E$ with respect to $\mathcal{F}$:
We say that $x_n\to\infty$ for a sequence in $E$, if for any finite subset $A\subset E$ there are only finite many $n\in\mathbb{N}$ with $x_n\in A$. Consider the set
$$
\xi_{\infty}:=\left\{(x_n)_{n\in\mathbb{N}}\in E: x_n\to\infty\text{ and }(f(x_n))_{n\in\mathbb{N}}\text{ converges for all }f\in\mathcal{F}\right\}.
$$
On $\xi_{\infty}$ we consider the following equivalence relation:
$$
(x_n)_{n\in\mathbb{N}}\sim (y_n)_{n\in\mathbb{N}}~\Leftrightarrow~\lim_{n\to\infty}f(x_n)=\lim_{n\to\infty}f(y_n)~~\forall f\in\mathcal{F}.
$$
Then it is
$$
\hat{E}\setminus E\simeq \xi_{\infty}/\sim.
$$

So my idea is that one maybe can show that
$$
\xi_{\infty}/\sim~\simeq~\mathbb{S}^1?
$$
Or maybe another way is better? Maybe something that starts like: Let $x\in\hat{E}\setminus E$, then there exists a sequence $(x_n)_{n\in\mathbb{N}}$ in $E$ with $x_n\to x$ as $n\to\infty$.
 A: First, $\mathcal F$ is needlessly complicated. Let $f_x = f(x,\cdot)$.
Since $|x| \le 1$, $f_x(y) \ge 0$ for all $y$.
If $y_n \to \infty$ then $y_n \neq x$ for almost all $n$, and since $|x|<1$ and $y_n$ is bounded, $f_x(y_n)$ can't converge to $0$ anyway.
Then you can see that $f_x(y_n)$ converges to $L$ if and only if $L > 0$ and $|x-y_n|$ converges to $\sqrt{(1-|x|^2)/L}$, so you might as well replace $f_x$ with the function $y \mapsto |x-y|$.
Next, if $(x_n) \in \xi_\infty$, then $(x_n)$ converges for the usual topology.  
Since $(x_n) \to \infty$, we must have $|x_n| \to 1$. Since $\overline{\Bbb D}$ is compact, if $(x_n)$ doesn't converge then it has two (or more) distinct adherence values, say $y_1$ and $y_2$ (necessarily on the unit circle). Then since $E$ is not entirely contained on the bissector line of $y_1$ and $y_2$ (it's not contained in any line passing through the origin) there are points $x$ in $E$ such that $|x-y_1| \neq |x-y_2|$. But then $f_x(x_n)$ can't converge.
So $(x_n) \in \xi_\infty \iff \lim x_n$ exists and is on the unit circle.
Finally, $(x_n) \sim (y_n) \iff \lim x_n = \lim y_n$ 
(if the limits are distinct, we can again find some point $x \in E$ such that $|x-\lim x_n| \neq |x - \lim y_n|$ which contradicts $(x_n) \sim (y_n)$).
So you have a well-defined injection $\xi_\infty/\sim \to S^1$, just by taking limits.
It is really not hard to see that this is surjective by density of the $2$-adic rationals. Finally you haven't said what is the topology on $\hat E$ so I'm not sure how to explain why this is a homeomorphism.
