Why is the set $A$ open? I was reading the proof of the theorem in the Principles of Mathematical Analysis written by Walter Rudin.
Something obfuscates me and I cannot figure it out. 
Here is the theorem (Theorem $8.5$ ):

Suppose the series $\sum a_n x^n$ and $\sum b_n x^n$ converge in the segment $S = (-R,R)$. Let $E$ be the set of all $x \in S$ at which 
  $$\sum_{n=0}^{\infty} a_n x^n=\sum_{n=0}^{\infty} b_n x^n$$ If $E$ has a limit point in $S$, then $a_n=b_n$ for $n = 0,1,2,....$ Hence the above equation holds for all $x \in S$.

In the proof, the author established a set named $A$ consisting all the limit points of $E$ in $S$. After that, he claimed $A$ is open.
In his proof about openess of $A$. He states:

If $x_0 \in A$, Theorem $8.4$ shows that $$f(x) = \sum_{n=0}^{\infty} d_n(x-x_0)^n (|x-x_0|<R-|x_0|)  \tag {22}$$

After that he proves $d_n =0$.

Thus $d_n = 0$ for all $n$, so that $f(x) =0 $ for all $x$ for which $(22)$ holds, i.e. in a neighborhood of $x_0$. This shows that $A$ is open, and completes the proof. 

However, my first question is that since $A$ contains only the limit points of $E$, could $A$ be a set that only contains one point? If so, $A$ would not be open.
Another question is that it only shows that there exists a neighborhood of $x_0$ in $S$, not one in $A$. Is there any more details about the link between the statements, or is his proof correct?

Of course, if $A$ had only a single point, it would not be an open set.
Rudin's argument shows that, if $x_0\in A$, then $A\supset\bigl(x_0-(R-|x_0|),x_0+(R-|x_0|\bigr)$, since he proved that both series have the same sum in this interval. Therefore, $A$ is an open set.
Perhaps that you are confused because of the use of limit points. You should keep in mind that if $E\supset(a,b)$, then the set of all limit points of $E$ also contains $(a,b)$.
