"Baby" Rudin theorem 8.5: is this proof correct? Theorem 8.5 of "Baby" Rudin is the following:

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 $n=0,1,2,\cdots$.

And here is my proof, which I want to get verified.

Put $c_n = a_n - b_n $, then $f(x) = \sum c_n x^n =0$ for all $x \in E $
Let $t$ be a limit point of $E$.
Then there is a sequence in $E$ which converges to $t$, and there exist a subsequence of it which consists of numbers bigger than $t$ or smaller than $t$.
wlog assume there is a increasing sequence $ \{ t_n \} $ which converges to $t$ and $t_n < t$ for all $n$
Then we have $f(t_n )=0$ for all $n$.
Since $f$ is continuous on $(-R, R)$, we have $f(t)=0$. (Note that $t \in S$, so $f(t)$ can be defined.)
Also, since $f$ is differentiable on $(-R, R)$, by mean value theorem, there exists a sequence $s_n $ such that $t_{n} < s_n < t_{n+1} $ and $f'(s_n )=0$ for all $n$.
It is obvious that $s_n$ converges to $t$, and since $f'$ is continuous on $(-R, R)$ too, so we have $f'(t)=0$.
We can repeat this, (rigorous proof can done by induction) to get $f^{(n)}(t)=0$ for all $n$.
Thus by Theorem 8.4 (which I state at the end of this question),
$f(x)=0$ for $x \in ( t-\epsilon , t+\epsilon ) \subset (-R, R) $, where $\epsilon >0$ is arbitrary.
Now we can repeat this stuff to get $f(x) =0$ for all $x \in (-R,R)$.
So $f^{(n)}(0)=0$ for all $n$, which gives $c_n =0$ for all $n$.

Here is Theorem 8.4, which I used above.

Suppose $f(x) = \sum_{n=0}^{\infty} c_{n} x^{n} $, the series converging in $|x|<R$. If $-R < a < R$, then $f$ can be expanded in a power series about the point $x=a$ which converges in $|x-a| < R-|a|$, and 
$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^{n} $

Thank you for reading my question.
Any commnets will help me a lot.
 A: A shorter variant of the proof that separates the computation of the Taylor/series coefficients at $x=t$ and determining their value is as follows:
You can expand the difference around $t$ so that
$$
\sum_{k=0}^\infty (a_k-b_k)x^k=\sum_{k=0}^\infty c_k(x-t)^k=c(x)
$$
for all $x$ with $|x-t|<r=R-|t|$.
Then as $c(t_k)=0$ and $c$ is continuous inside its radius of convergence, one finds $c_0=c(t)=0$. For $t_n\ne t$ we still have $c(t_n)/(t-t_n)=0$, so that the series
$\sum_{k=0}^\infty c_{k+1}(x-t)^k$ has the same properties as $c$, so that $c_1=0$ follows etc. Per induction $c_k=0$ for all $k$.
Shifting the expansion point back, resp. inverting the original shift to $t$ finds that $a_k-b_k=0$ for all $k$. Or apply the ironing argument of your proof in that the zero set of $f$ now includes the full interval $(t-r,t+r)$, which can now be expanded towards zero by repeating the proof at the endpoint closer to zero until zero is contained and thus the expansion at zero has all coefficients zero. 

The same argument, more compact: If $f\ne 0$, then the root at $x=t$ has a finite multiplicity $m$. Then one can split off the linear factor $x-t$ $m$ times to get $f(t)=(x-t)^mg(t)$ where $g$ is still a continuous function and $g(t)\ne 0$. But $g(t_n)=\frac{f(t_n)}{(t_k-t)^m}=0$, which is a contradiction to continuity in $g(t)=\lim_{n\to\infty}g(t_n)$.
A: @sansae
I don't think yours is a valid one.
Because you constructed a sequence only in E which may be a set of discrete points where continuity of f cannot be applied.
So, I'd like to suggest you that you should construct a sequence in S first and you should also show that the set of limit points of E is a subset of E first. Then only you can apply your procedure.
