Uniform Differentiability implies $f$ is $C^1$ This exercise problem in Abbott's text asks, for the proof of the fact: if $f$ is uniformly differentiable on $A$, then $f$ is $C^1$ on $A$. I'd like someone to verify if (a) my proof is correct (b) my proof is readable or gibberish.

[Abbott, 6.2.4] Review exercise 5.2.8 which includes the definition for a uniformly differentiable function. Use the results discussed in section 6.2 to show that if $\displaystyle f$ is uniformly differentiable, then $\displaystyle f'$ is continuous.

Proof.
Suppose $\displaystyle f$ is uniformly differentiable on $\displaystyle A$. Recall from the definition, that for all $\displaystyle \epsilon  >0$, there exists $\displaystyle \delta ( \epsilon )  >0$, such that for all $\displaystyle x,y\in A$ satisfying $\displaystyle |x-y|< \delta $, the distance $\displaystyle \left| \frac{f( x) -f( y)}{x-y} -f'( y)\right| < \epsilon $.
Pick an arbitrary $\displaystyle \epsilon  >0$.
Let $\displaystyle c\in A$ be a fixed arbitrary point and suppose $\displaystyle ( x_{n}) \subseteq A$ with $\displaystyle x_{n} \neq c$ for all $\displaystyle n\in N$ be an arbitrary sequence that converges to $\displaystyle c$. So, there exists $\displaystyle N( \delta ) \in \mathbf{N}$, such that $\displaystyle |x_{n} -c|< \delta $ for all $\displaystyle n\geq N$. But, if the distance $\displaystyle |x_{n} -c|< \delta $, it follows that
\begin{equation*}
\left| \frac{f( x_{n}) -f( c)}{x_{n} -c} -f'( c)\right| < \epsilon 
\end{equation*}
for all $\displaystyle n\geq N$.
Since the above is true for all $\displaystyle c\in A$, the sequence of functions $\displaystyle ( d_{n})$ given by
\begin{equation*}
d_{n}( y) =\begin{cases}
\frac{f( y) -f( x_{n})}{y-x_{n}} & \text{if } y\neq x_{n}\\
f'( y) & \text{if } y=x_{n}
\end{cases}
\end{equation*}
converge pointwise to $\displaystyle f'( y)$ for all $\displaystyle y\in A$ and $\displaystyle n\geq N$.
As $\displaystyle N$ is only a function of $\displaystyle \epsilon $, independent of $\displaystyle y$, $\displaystyle ( d_{n}( y))$ converges uniformly to the limit function $\displaystyle f'( y)$.

We can also prove that each $\displaystyle d_{n}$ is continuous.
Suppose $\displaystyle y\rightarrow l$ and $\displaystyle l\neq x_{n}$, then both the numerator and denominator are convergent sequences and $\displaystyle \lim _{y\rightarrow l} d_{n}( y) =\frac{f( l) -f( x_{n})}{l-x_{n}} =d_{n}( l)$ by the Algebraic limit theorem and the fact that $\displaystyle f$ is continuous.
If $\displaystyle l=x_{n}$, then $\displaystyle \lim _{y\rightarrow x_{n}} d_{n}( y) =\lim _{y\rightarrow x_{n}}\frac{f( y) -f( x_{n})}{y-x_{n}} =f'( x_{n}) =d_{n}( x_{n})$ by definition.
So, each $\displaystyle d_{n}$ is continuous on $\displaystyle A$.
By the continuous limit theorem, $\displaystyle \lim d_{n} =f'$ is continuous on $\displaystyle A$.
 A: Posting an answer, using the suggestion from the comments.
Uniform convergence

Define the sequence of functions $\displaystyle ( d_{n})$ given by
\begin{equation*}
d_{n}( y) =\frac{f\left( y+\frac{1}{n}\right) -f( y)}{\left( y+\frac{1}{n}\right) -y}
\end{equation*}
Let $\displaystyle \epsilon  >0$ be arbitrary. Since $\displaystyle f$ is uniformly differentiable, there exists $\displaystyle \delta  >0,$ such that for $\displaystyle y\in A$, if $\displaystyle \frac{1}{n} < \delta $, it follows that $\displaystyle \left| \frac{f\left( y+\frac{1}{n}\right) -f( y)}{\left( y+\frac{1}{n}\right) -y} -f'( y)\right| < \epsilon $. By the Archimedean property, there exists $\displaystyle N\in \mathbf{N}$, such that $\displaystyle \frac{1}{N} < \delta $. So, there exists $\displaystyle N$, such that $\displaystyle | d_{n}( y) -f'( y)| < \epsilon $ for all $\displaystyle n\geq N$. But, as $\displaystyle \epsilon $ was arbitrary, this is true for all $\displaystyle \epsilon  >0$. So, $\displaystyle ( d_{n})$ uniformly converges to $\displaystyle f'$ on $\displaystyle A$.
Each $\displaystyle d_{n}$ is continuous.

We can also prove that each $\displaystyle d_{n}$ is continuous.
Let $\displaystyle c$ be an arbitrary point in $\displaystyle A$.
$\displaystyle \begin{array}{ c l l }
\lim _{y\rightarrow c} d_{n}( y) & =\lim _{y\rightarrow c}\frac{f\left( y+\frac{1}{n}\right) -f( y)}{\left( y+\frac{1}{n}\right) -y} & \\
 & =\lim _{y\rightarrow c} n\cdot \left\{f\left( y+\frac{1}{n}\right) -f( y)\right\} & \\
 & =n\cdot \left\{\lim _{y\rightarrow c} f\left( y+\frac{1}{n}\right) -\lim _{y\rightarrow c} f( y)\right\} & \left\{\text{Algebraic Limit Theorem}\right\}\\
 & n\cdot \left\{f\left( c+\frac{1}{n}\right) -f( c)\right\} &  \begin{array}{{>{\displaystyle}l}}
\{\text{Since } y+\frac{1}{n} \ \text{ and } f\ \text{are continuous,}\\
\text{the composition } f( y+1/n) \ \\
\text{is continuous}\}
\end{array}\\
 & =d_{n}( c) & 
\end{array}$
Thus, each $\displaystyle d_{n}$ is continuous.
By the continuous limit theorem, $\displaystyle f'$ is continuous on $\displaystyle A$.
