Alternative Answer for Baby Rudin $4.1$: Does $\lim_{h\rightarrow 0}[f(x+h)-f(x-h)]=0$ imply continuity? Can anyone point out the mistake I am making for the following question?
The problem is written in Baby Rudin Chapter $4$.

Suppose $f$ is a real function defined on $E$, which satisfies,
$\lim_{h\rightarrow 0}[f(x+h)-f(x-h)]=0$
for every $x\in R^1$. Does this imply that $f$ is continuous?

My solution for this problem is of following.

From condition given in the question,
$\lim_{h\rightarrow 0}[f(x+h)-f(x)]=\lim_{h\rightarrow 0}[f(x-h)-f(x)]$
As a way of contradiction, suppose that $f$ is not continuous.
Therefore, There exists a  $x\in R$ such that
$\exists \epsilon>0 ,s.t.$ $d(f(x^\prime),f(x))>\epsilon$ for all $x^\prime \in B_\delta(x), \forall \delta>0$
Generate a decreasing sequence $\{\delta_n\}_{n=1}^\infty$ such that $\delta_k>\delta_{k-1}, \forall k$.
By taking out a $x_{n}^{\prime}$ from each open ball defined around $x$ with radius $\delta_{n}$, we can create a partial sequence $\{x_{n_k}\}$ that converges to $x$ but the mapping $f(x^\prime)$ does not.
This contradicts the assumption, since it requires that each of the limits of functions on both sides to converge.
$\lim_{h\rightarrow 0}[f(x+h)-f(x)]=\lim_{h\rightarrow 0}[f(x-h)-f(x)]$

There is a solution manual (https://minds.wisconsin.edu/bitstream/handle/1793/67009/rudin%20ch%204.pdf?sequence=8&isAllowed=y) which gives an explanation of a counterexample, but I cant figure out which part in my proof is wrong.
 A: Bungo beat me to this in the comments.
I am just expanding upon their answer.
The first claim you make in your proof is incorrect.
To convince yourself, consider 
$$
f(x)=\begin{cases}
1/|x| & \text{if }x \neq 0 \\
0     & \text{if }x =    0.
\end{cases}
$$
In general, $\lim_{n}(a_{n}-b_{n})=0$ does not imply $\lim_{n}a_{n}=\lim_{n}b_{n}$.
The converse, however, is true from the sum law of limits.
A: There are several mistakes in your work. Two of them are not serious; they can be side-stepped in this particular proof but they are important things to list as they concern correct use of definitions and correct proof-technique:
Small Errors:
$\underline{\textit{You did not negate continuity correctly:}}$

$\exists \epsilon>0 ,s.t.$ $d(f(x^\prime),f(x))>\epsilon$ for all $x^\prime \in B_\delta(x), \forall \delta>0$

That is not exactly the negation of continuity at $x$. This is:

$\exists \epsilon>0,\ s.t. \forall \delta > 0 \underline{\textbf{ there exists }} x' \in B_\delta(x)\ s.t. d(f(x'), f(x)) > \epsilon$

$\underline{\textit{You probably want the sequence of $\delta_n$ to approach $0$:}}$

By taking out a $x_{n}^{\prime}$ from each open ball defined around $x$ with radius $\delta_{n}$, we can create a partial sequence $\{x_{n_k}\}$ that converges to $x$ but the mapping $f(x^\prime)$ does not.

You can have a sequence of decreasing $\delta_n$ that does not approach $0$. In that case, the sequence elements $x_n'$ you picked out from each $B_{\delta_n}(x)$ may not even converge to $x$ as you wanted them to.
Okay, now let us look at the serious errors that break your proof:
Serious Errors:
$\underline{\textit{Your very first step is wrong:}}$

From condition given in the question,
$\lim_{h\rightarrow 0}[f(x+h)-f(x)]=\lim_{h\rightarrow 0}[f(x-h)-f(x)]$

No, you can have functions that satisfy $$\lim_{h\rightarrow 0}[f(x+h)-f(x-h)] = 0$$ but neither $$RL\ (\text{Right Limit}) = \lim_{h\rightarrow 0}[f(x+h)-f(x)]$$ nor $$LL \ (\text{Left Limit}) = \lim_{h\rightarrow 0}[f(x-h)-f(x)]$$ exist, let alone equal one another.
E.g. pick your favorite even function with an asymptote at $0$ like $f(x) = \frac{1}{x^2}$ or $f(x) = \frac{1}{x^4}$, etc and define $f(0)$ to be whatever real you like. You can check that neither the $RL$ nor $LL$ exist at $x = 0$ due to the asymptote. But because the function is even, $f(0+h) - f(0-h) = 0$ so that $\lim_{h\rightarrow 0}[f(0+h)-f(0-h)] = 0$.
$\underline{\textit{And even if your first step was correct:}}$
In your defense, I will admit that the example given in your link:
$$
f(x) = \begin{cases}1 &\text{ $x$ an integer} \\
0 &\text{ otherwise }\end{cases}
$$ does satisfy both $$\lim_{h\rightarrow 0}[f(x+h)-f(x-h)]=0$$ and $$\lim_{h\rightarrow 0}[f(x+h)-f(x)]=\lim_{h\rightarrow 0}[f(x-h)-f(x)]$$ But even then:

This contradicts the assumption, since it requires that each of the limits of functions on both sides to converge.
$\lim_{h\rightarrow 0}[f(x+h)-f(x)]=\lim_{h\rightarrow 0}[f(x-h)-f(x)]$

No it doesn't contradict that assumption. $\lim_{h\rightarrow 0}[f(x+h)-f(x)]$ being equal to $\lim_{h\rightarrow 0}[f(x-h)-f(x)]$ does not mean that either of those limits should converge to $0$.
Indeed, for the example given in the question, you can check that at integer points $x$,
$\lim_{h\rightarrow 0}[f(x+h)-f(x)] = \lim_{h\rightarrow 0}(0 - 1) = -1$ and
$\lim_{h\rightarrow 0}[f(x-h)-f(x)] = \lim_{h\rightarrow 0}(0 - 1) = -1$.
Hence, the two limits are equal to each other but neither of them are $0$. So, the fact that your picked out sequence $x_n'$ converges to $x$ while the $f(x_n')$'s do not converge to $f(x)$ is not a contradiction.
A: Apart from the problem pointed out by Bungo, even if the limits do exist, they are non necessarily zero.  Thus, there is no contradiction between the conditions that $\ \lim_\limits{k\rightarrow\infty} x_{n_k}=x\ $, $\ \lim_\limits{k\rightarrow\infty} f\left(x_{n_k}\right) \ne f\left(x\right)\ $, and $\lim_{h\rightarrow 0}[f(x+h)-f(x)]=\lim_{h\rightarrow 0}[f(x-h)-f(x)]\ $.  We could have
$$
\lim_{h\rightarrow 0}[f(x+h)-f(x)]=\lim_{h\rightarrow 0}[f(x-h)-f(x)]= C\ne0\ \ \mbox{and}\\
\lim_\limits{k\rightarrow\infty} f\left(x_{n_k}\right) = f\left(x\right) + C\ ,
$$
for instance.
A: Thank you for all the comments and answers!
I guess I have not quite yet been used to properties of summation of limits.
Example : $\lim_{x\rightarrow0}(f(x)+g(x))$ is not necessary $\lim f(x)+\lim g(x)$ .
