Theorem 7.18 in PMA Rudin: Existence of an everywhere continuous but nowhere differentiable real function on the real line Here is Theorem 7.18 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition: 

There exists a real continuous function on the real line which is nowhere differentiable. 

And, here is Rudin's proof ( steps wherein I've been unable to figure out on my own and hence would appreciate the help of the Math SE community): 

Define 
  $$\tag{34} \varphi(x) = \lvert x \rvert \qquad \qquad (-1 \leq x \leq 1) $$
  and extend the definition of $\varphi(x)$ to all real $x$ by requiring that 
  $$ \tag{35} \varphi(x+2) = \varphi(x). $$
  Then, for all $s$ and $t$, 
  $$\tag{36}  \lvert \varphi(s) - \varphi(t) \rvert \leq \lvert s-t \rvert. $$
  [ How to obtain the inequality in (36)? ] 
  In particular, $\varphi$ is continuous on $\mathbb{R}^1$. Define 
  $$ \tag{37} f(x) = \sum_{n=0}^\infty \left( \frac{3}{4} \right)^n \varphi \left( 4^n x \right). $$
  Since $0 \leq \varphi \leq 1$, Theorem 7.10 shows that the series (37) converges uniformly on $\mathbb{R}^1$. By Theorem 7.12, $f$ is continuous on $\mathbb{R}^1$. 
Now fix a real number $x$ and a positive integer $m$. Put 
  $$ \tag{38} \delta_m = \pm \frac{1}{2} \cdot 4^{-m} $$
  where the sign is so chosen that no integer lies between $4^m x$ and $4^m \left( x + \delta_m \right)$. This can be done, since $4^m \left\lvert \delta_m \right\rvert = \frac{1}{2}$. Define 
  $$ \tag{39} \gamma_n = \frac{ \varphi \left( 4^n \left( x + \delta_m \right)  \right) - \varphi \left( 4^n x \right)  }{ \delta_m }. $$
  When $n > m$, then $4^n \delta_m$ is an even integer, so that $\gamma_n = 0$. When $0 \leq n \leq m$, (36) implies that $\left\lvert \gamma_n \right\rvert \leq 4^n$. 
Since $\left\lvert \gamma_m \right\rvert = 4^m$ [ How? ], we conclude that 
  $$
\begin{align}
\left\lvert \frac{ f \left( x + \delta_m \right) - f(x)  }{ \delta_m  }  \right\rvert &= \left\lvert \sum_{n=0}^m \left( \frac{3}{4} \right)^n \gamma_n  \right\rvert \\ 
&\geq 3^m - \sum_{n=0}^{m-1} 3^n \\
&= \frac{1}{2} \left( 3^m + 1 \right).
\end{align}
$$
  As $m \to \infty$, $\gamma_m \to 0$. It follows that $f$ is not differentiable at $x$. 

Here is Theorem 7.10 in Baby Rudin, 3rd edition: 

Suppose $\left\{ f_n \right\}$ is a sequence of functions defined on $E$, and suppose $$ \left\lvert f_n (x) \right\rvert \leq M_n \qquad \qquad (x \in E, \ n = 1, 2, 3, \ldots \ ). $$
  Then $ \sum f_n $ converges uniformly on $E$ if $ \sum M_n$ converges. 
Note that the converse is not asserted ( and is, in fact, not true). 

And, here is Theorem 7.12: 

If $\left\{ f_n \right\}$ is a sequence of continuous functions on $E$, and if $f_n \to f$ uniformly on $E$, then $f$ is continuous on $E$. 

The rest of the proof I understand, I think. 
However, I would appreciate if someone could give the crux of the procedure involved in the construction of this particular example and also give a blueprint for constructing this class of functions. 
 A: Note that the the sign is so chosen that no integer lies between $4^m x$ and $4^m \left( x + \delta_m \right)$. Because of this, and $4^n \delta_m = \pm \frac{1}{2}$, we have $\left\lvert \varphi \left( 4^n \left( x + \delta_m \right)  \right) - \varphi \left( 4^n x \right) \right\rvert = \frac{1}{2} $. Thus, $\left\lvert \gamma_m \right\rvert = 4^m$
.
A: The blueprint is: create a limit of function that oscillate more and more** on small scales, but with higher-frequency oscillations being damped quickly ($(\frac{3}{4})^n$ factor) [1].
** The $4^n$ factor inside $\varphi(4^nx)$
[1] - https://ocw.mit.edu/courses/mathematics/18-100b-analysis-i-fall-2010/readings-notes/MIT18_100BF10_WeierstFunc.pdf
A: Due to $(35)$, $φ$ is periodic with period $2$.
$\tag{36}|φ(s)−φ(t)|= ||s|−|t||≤|s−t|.$
Due to $(36)$, $φ$ is continuous on $\mathbb{R}^1$.
Since,
$$ \left|\left( \frac{3}{4} \right)^n \varphi \left( 4^n x \right)\right|\le \left( \frac{3}{4} \right)^n$$
and $\sum \left( \frac{3}{4} \right)^n$ converges, Theorem 7.10 shows that the series $(37)$ converges uniformly. Since the limit function $f$ is the series of continuous functions on $\mathbb{R}^1$, by Theorem 7.12, $f$ is continuous on $\mathbb{R}^1$ too.
The rest is the proof for nowhere differentiability of $f$.
Indeed, the series $(37)$ is made up of terms whose wavelength and amplitude gradually tends to zero each. Since the wavelengths become gradually decrease, the slopes of the corresponding segments of the wave gradually increase so that the differentiability of the segments gradually loses. This is so arranged for the series to converge uniformly to a nowhere differentiable continuous function on $\mathbb{R}^1$.
The followings present the first four terms of the series and their sum which looks like a sharp zigzag.


