Proof of Banach's contraction mapping theorem I am self-learning Real Analysis from Stephen Abbott's, Understanding Analysis. Exercise 4.3.11 asks to prove the Contraction Mapping Theorem, which is extremely important for fixed-point iteration to numerically solve for the zeroes of an equation, and I think it even extends to PDEs, where a function is a fixed point in infinite dimensional function spaces.
The text provides several hints in this regard. I'd like to ask for some help, in (a) verifying whether my proof steps are rigorous and sufficient (b) some more hints for the other parts.

Contraction Mapping Theorem. Let $\displaystyle f$ be a function defined on all of $\displaystyle \mathbf{R}$, and assume there is a constant $\displaystyle c$ such that $\displaystyle 0< c< 1$ and
\begin{equation*}
| f( x) -f( y)| \leq c| x-y| 
\end{equation*}
(a) Show that $\displaystyle f$ is continuous on $\displaystyle \mathbf{R}$.
(b) Pick some point $y_1 \in \mathbf{R}$ and construct the sequence
$$(y_1,f(y_1),f(f(y_1)),\ldots)$$
In general, if $y_{n+1}=f(y_n)$, show that the resulting sequence $(y_n)$ is a Cauchy sequence. Hence, we may let $y = \lim y_n$.
(c) Prove that $y$ is a fixed point of $f$ (i.e. $f(y) = y$) and that it is unique in this regard.
(d) Finally, prove that if $x$ is any arbitrary point in $\mathbf{R}$, then the sequence $(x,f(x),f(f(x),\ldots)$ converges to $y$ defined in (b).

Proof.
(a) Let $\displaystyle x_{0} \in \mathbf{R}$ and let $\displaystyle ( x_{n})\rightarrow x_{0}$ be an arbitrary sequence. By definition, for all $\displaystyle \delta  >0$, there exists a $\displaystyle N\in \mathbf{N}$, such that $\displaystyle | x_{n} -x_{0}| < \delta $ for all $\displaystyle n\geq N$.
We are interested to make the distance $\displaystyle | f( x) -f( x_{0})| < \epsilon $. Replacing $\displaystyle | f( x) -f( x_{0})| $ by $\displaystyle c| x-x_{0}| $ strengthens the condition we want to prove. So, we have $\displaystyle | x-x_{0}| < \frac{\epsilon }{c}$.
Pick $\displaystyle \delta =\frac{\epsilon }{c}$. Thus, for all $\displaystyle \epsilon  >0$, there exists $\displaystyle \delta  >0$, such that for all $\displaystyle | x-x_{0}| < \delta $, we have $\displaystyle | f( x) -f( x_{0})| < \epsilon $.
Consequently, $\displaystyle f( x)$ is continuous on $\displaystyle \mathbf{R}$.
(b) We are interested to make the distance $\displaystyle | y_{n+m} -y_{m}| $ as small as we please. Pick an arbitrary $\displaystyle \epsilon  >0$. Let's explore the expression $\displaystyle | y_{n+m} -y_{m}| $.
\begin{equation*}
\begin{array}{ r l c }
| y_{n+m} -y_{m}|  & =| y_{n+m} -y_{n+m-1} +y_{n+m-1} -y_{n+m-2} +\dotsc +y_{m+1} -y_{m}|  & \\
 & \leq | y_{n+m} -y_{n+m-1} |+|y_{n+m-1} -y_{n+m-2} |+\dotsc +|y_{m+1} -y_{m}|  & \left\{\text{Triangle Inequality}\right\}\\
 & \leq c| y_{n+m-1} -y_{n+m-2}| +c|y_{n+m-2} -y_{n+m-3} |+\dotsc  & \\
 & +\ c| y_{m} -y_{m-1}|  & \\
 & \leq \left( c^{n+m-2} +\dotsc +c^{m-1}\right)| y_{2} -y_{1}|  & 
\end{array}
\end{equation*}
If $\displaystyle 0< b< 1$, we know that the sequence $\displaystyle \left( b^{n}\right)\rightarrow 0$. Therefore, for all $\displaystyle \epsilon  >0$, there exists $\displaystyle N >0$, such that the entire expression $\displaystyle \left( c^{n+m-2} +\dotsc +c^{m-1}\right) < \epsilon /| y_{2} -y_{1}| $ for all $\displaystyle n >m\geq N$. Consequently, $\displaystyle | y_{n+m} -y_{m}| < \epsilon $ for all $\displaystyle n >m\geq N$. Thus, $\displaystyle ( y_{n})$ is Cauchy. Cauchy sequences are convergent, so let $\displaystyle \lim y_{n} =y$.
(c) $\displaystyle y$ is a limit point of domain of $\displaystyle f$, since there exists a sequence $\displaystyle ( y_{n})$ in $\displaystyle \mathbf{R}$, such that $\displaystyle ( y_{n})\rightarrow y$, with $\displaystyle y_{n} \neq y$ for all $\displaystyle n\in \mathbf{N}$.
By the definition of functional continuity, since $\displaystyle f$ is continous, if $\displaystyle x$ is a limit point of the domain of $\displaystyle f$, for all sequences $\displaystyle ( x_{n})\rightarrow x$, it follows that $\displaystyle f( x_{n})\rightarrow f( x)$.
Consequently, as $\displaystyle ( y_{n})\rightarrow y$, $\displaystyle \lim f( y_{n}) =f( y)$. But, $\displaystyle \lim f( y_{n}) =\lim y_{n+1} =y$. Thus, $\displaystyle f( y) =y$. $\displaystyle y$ is a fixed point of $\displaystyle f$.
Assume that $\displaystyle y$ and $\displaystyle y'$ are points such that $\displaystyle f( y) =y$ and $\displaystyle f( y') =y'$. We have that, $\displaystyle | f( y) -f( y')| =| y-y'| $. But, $\displaystyle f$ is a contraction mapping. So, $\displaystyle | f( y) -f( y')| \leq c| y-y'| $. Thus,
$\displaystyle ( 1-c)| y-y'| \leq 0$. But, $\displaystyle 1 >c\Longrightarrow 1-c >0$. So, the only possibility is $\displaystyle | y\ -\ y'| =0$. It follows that $\displaystyle y=y'$.
(d) As seen earlier, the sequence $\displaystyle ( x,f( x) ,f( f( x)) ,\dotsc )$ \ converges to some $\displaystyle x'$ such that $\displaystyle f( x') =x'$. But, the fixed points of $\displaystyle f$ are unique. So, $\displaystyle x'=y$.
 A: 
For the sake of completeness, posting this as an answer.

(b) We are interested to make the distance $\displaystyle | y_{n} -y_{m}| $ as small as we please. Pick an arbitrary $\displaystyle \epsilon  >0$. Let's explore the expression $\displaystyle | y_{n} -y_{m}| $.
\begin{equation*}
\begin{array}{ r l c }
| y_{n} -y_{m}|  & =| y_{n} -y_{n-1} +y_{n-1} -y_{n-2} +\dotsc +y_{m+1} -y_{m}|  & \\
 & \leq | y_{n} -y_{n-1} |+|y_{n-1} -y_{n-2} |+\dotsc +|y_{m+1} -y_{m}|  & \left\{\text{Triangle Inequality}\right\}\\
 & \leq c| y_{n-1} -y_{n-2}| +c|y_{n-2} -y_{n-3} |+\dotsc  & \\
 & +\ c| y_{m} -y_{m-1}|  & \left\{f\ \text{is a contraction map}\right\}\\
 & \leq \left( c^{n-2} +\dotsc +c^{m-1}\right)| y_{2} -y_{1}|  & \\
 & =c^{m-1}\left( c^{n-m-1} +\dotsc +1\right) |y_{2} -y_{1} | & \\
 & =c^{m-1}\left(\frac{1-c^{n-m}}{1-c}\right) |y_{2} -y_{1} | & \left\{\text{Assuming } n >m\right\}\\
 & < c^{m-1}\left(\frac{1}{1-c}\right)| y_{2} -y_{1}|  & 
\end{array}
\end{equation*}
If $\displaystyle 0< b< 1$, we know that the sequence $\displaystyle \left( b^{n}\right)\rightarrow 0$. Therefore, for all $\displaystyle \epsilon  >0$, if we pick $\displaystyle N >\frac{\log \epsilon }{\log b}$, then $\displaystyle b^{n} < \epsilon $ for all $\displaystyle n\geq N$.
Consequently, if we pick a large $\displaystyle N$ such that
\begin{equation*}
c^{N} < \epsilon \ \left(\frac{1-c}{| y_{2} -y_{1}| }\right)
\end{equation*}
then for all $\displaystyle n >m >N$, we have:
\begin{equation*}
| y_{n} -y_{m}| < c^{m-1}\left(\frac{1}{1-c}\right)| y_{2} -y_{1}| < \epsilon \ \left(\frac{1-c}{| y_{2} -y_{1}| }\right) \cdot \left(\frac{1}{1-c}\right)| y_{2} -y_{1}| =\epsilon 
\end{equation*}
Thus, $\displaystyle ( y_{n})$ is Cauchy. Cauchy sequences are convergent, so let $\displaystyle \lim y_{n} =y$.
