I am sorry if the title of this post is confusing, feel free to edit it. My question is as follows:
I am working on an exercise stated as follows:
Let $\mathcal{E}$ be the set of holomorphic functions $f(z)$ defined as power series $f(z)=\sum_{n=1}^{\infty}a_{n}z^{n}$ with radius of convergence equal to $1$. Denoting $C$ to the unit circle, and for $f\in\mathcal{E}$, we define $$S(f):=\{z\in C:\ f\ \text{cannot be extended as a holomorphic function in a neighborhood of}\ z\}.$$
Show that $S(f)$ is closed and non-empty.
For $z_{0}\in C$, find an example such that $S(f)=\{z_{0}\}$.
I found a similar post here: Holomorphic function $f$ in $D$ which is not holomorphically extendible across any point in $\partial D$.
I tried to mimic the proof in the above post. As $\overline{\mathbb{D}}$ is sequentially compact, for any $z\in C$, there exists a sequence $\{z_{n}\}_{n=1}^{\infty}$ in $\mathbb{D}$ that converges to $z$. We know recall
[Weierstrass Theorem on Zeros of Holomorphic Functions] Assume $V$ is a proper open set of the Riemann Sphere $\mathbb{C}_{\infty}$, and let $A\subset V$ be a set that has no accumulation point in $V$. Then, there exists a function $f$ holomorphic on $V$ such that the set of zeros $Z(f)$ is exactly $A$, counting the multiplicities.
In our case, $V=\mathbb{D}$ and $A=\{z_{n}\}_{n=1}^{\infty}$. Note that the limit point of $A$ is on $\partial D$, so $A$ has no accumulation point in $\mathbb{D}$. Hence, by the theorem, there exists a holomorphic function $g$ in $\mathbb{D}$ such that $g(z_{n})=0$ for all $z_{n}\in A$.
But then I don't know how to proceed. The original post said that since the zeros are isolated, then we cannot extend beyond $z\in C$. I don't quite follow this argument.
Also, my proof seems backward... As saulspatz suggested, $f$ is given, and I need to show that there exists $z\in C$ such that $f$ cannot be analytically continued.
I don't quite know how to prove the set is closed.
I also don't really know how to find the example for $2$. I took a look at https://mathoverflow.net/questions/10831/example-of-continuous-function-that-is-analytic-on-the-interior-but-cannot-be-an, but did not find a proper example.
Thanks for any help!
Edit 1: Proof of Q1 part 1
Thanks to Conrad's suggestion, we show $C\setminus S(f)$ open. To show this, we show that for any $z\in C\setminus S(f)$, there exists $r=r(z)>0$ such that $D(z,r)\cap C\subset C\setminus S(f)$.
Indeed, let $z\in C\setminus S(f)$, i.e. $f$ can be analytically continued around $z$. Hence, there exists $r=r(z)>0$ such that there exists a holomorphic function $g$ on $D(z,r)$ such that $g=f$ on $D(z,r)\cap D$.
Let $z_{0}\in D(z,r)\cap C$, take $r_{0}>0$ such that $D(z_{0}, r_{0})\subsetneq D(z,r)$. Then, it is clear that $g$ is holomorphic on $D(z_{0}, r_{0})$ and $f=g$ on $D(z_{0}, r_{0})\cap\mathbb{D}\cap D(z,r)\cap\mathbb{D}$.
Hence, $f$ has an analytic continuation around $z_{0}$. This means that $z_{0}\in C\setminus S(f)\implies D(z,r)\cap C\subset C\setminus S(f)$.
We are done
Edit 2: Example for Q2.
Taking $z_{0}\in C$, i.e. $|z_{0}|=1$ and consider the series $$f(z):=\sum_{n=0}^{\infty}\Bigg(\dfrac{1}{z_{0}}\Bigg)^{n+1}z^{n}.$$ The radius of convergence is $$R:=\limsup_{n\rightarrow\infty}\Bigg|\dfrac{1}{z_{0}}\Bigg|^{\frac{n+1}{n}}=1.$$
So in $\mathbb{D}$, we have $$f(z)=\dfrac{1}{z_{0}}\sum_{n=0}^{\infty}\Bigg(\dfrac{z}{z_{0}}\Bigg)^{n}=\dfrac{1}{z_{0}-z}.$$
Note that the function $h(z):=\frac{1}{z_{0}-z}$ is holomorphic everywhere on $\overline{\mathbb{D}}$ except at $z_{0}$.
But $h(z)$ is not bounded around $z_{0}$ so by Riemann's theorem on removable singularity that $z_{0}$ is not removable.
Hence, we have $g$ analytic on $\overline{\mathbb{D}}\setminus\{z_{0}\}$ and $f=g$ on $\mathbb{D}$, and thus $f$ has an analytic continuation from $\mathbb{D}$ to $C$, except at the point $z_{0}$.
$S(f)=\{z_{0}\}$.
Edit 3:
Okay I proved that $S(f)$ is non-empty, see my own answer of the question below.
Okay, for other proofs, please see the edit of my post. I am answering my own question to prove that $S(f)\neq\varnothing$.
Suppose $S(f)=\varnothing$, then all the points on $C$ can be analytically continued to for $f$. So for each $z\in C$, we have $D(z,r)$ for $f$ to have the analytic continuation.
Then, as $\overline{\mathbb{D}}$ is compact, the covering $\Big(\bigcup_{z\in C}D(z,r)\Big)\cup\mathbb{D}$ gives us finite sub cover $\{D(z_{i}, r_{i})\}_{i=1}^{N}$.
We define $\epsilon:=\min_{i}r_{i}$, and note that this finite sub-cover will cover the annulus $D(0,1+\epsilon)\setminus\mathbb{D}$.
Define $f_{i}$ to be the corresponding analytic continuation of $f$ to $D(z_{i}, r_{i})$. Then, for each $z$ in this annulus, we define $$g(z)=f_{i}(z)$$ if $z\in D(z_{i}, r_{i})$. This is well-defined since every point in the annulus is in at least one covering ball, and in the intersection of two balls, $f_{i}(z)=f=f_{j}(z)$ by the analytic continuation.
Now, define $h(z)=f(z)$ for $z\in\mathbb{D}$ and $h(z)=g(z)$ for $z$ in annulus.
It follows that $h(z)$ is holomorphic in $D(0,1+c)$ for some $c<\epsilon$. So it has a power expansion $\sum_{n=1}^{\infty}b_{n}z^{n}$, but this will also represent $f(z)=\sum_{n=1}^{\infty}a_{n}z^{n}$ inside $\mathbb{D}$, this forces $a_{n}=b_{n}$, so $f(z)$ has radius of convergence $1+c>1$, a contradiction.