To solve question 23 you use the inequalities from questions 21 and 22, which I give here with their proofs.
Qu. 21) If
$$|x-x_0|<\min\bigg(\frac{\varepsilon}{2(|y_0|+1)}, 1 \bigg)\quad\text{and}\quad |y-y_0|< \frac{\varepsilon}{2(|x_0|+1)} \tag{1}$$
then $|xy-x_0y_0|<\varepsilon$.
Proof:
Since $|x-x_0|<1$ we have $|x|-|x_0|<|x-x_0|<1\Longrightarrow |x|<|x_0|+1$, and so
\begin{align*}
|xy-x_0y_0|&=|x(y-y_0)-y_0(x_0-x)|\\
&\le|x||y-y_0|+|y_0||x_0-x|\\
&< (1+ |x_0|)\frac{\varepsilon}{2(|x_0|+1)}+|y_0|\frac{\varepsilon}{2(|y_0|+1)}\\
&<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon.
\end{align*}
Qu. 22) If $y_0\neq0$ and
$$|y-y_0|<\min\bigg(\frac{|y_0|}{2}, \frac{\varepsilon |y_0|^2}{2} \bigg), \tag{2}$$
then $y\neq0$ and
$$\bigg| \frac{1}{y}-\frac{1}{y_0}\bigg|<\varepsilon.$$
Proof:
$|y_0|-|y|\le|y-y_0|<\frac{|y_0|}{2}\Longrightarrow |y|>\frac{|y_0|}{2}$. Since $y\neq0$ we have
$$\frac{1}{|y|}<\frac{2}{|y_0|}.$$
Then
$$\bigg| \frac{1}{y}-\frac{1}{y_0}\bigg|=\frac{|y-y_0|}{|y||y_0|}<\frac{1}{|y|}\cdot\frac{2}{|y_0|}\cdot\frac{\varepsilon |y_0|^2}{2}=\varepsilon.$$
Qu. 23) If $y_0\neq 0$ and
$$|y-y_0|<\min\bigg( \frac{|y_0|}{2},\frac{\varepsilon|y_0|^2}{4(|x_0|+1)} \bigg) \qquad\text{and}\qquad |x-x_0|<\min\bigg(\frac{\varepsilon}{2(1/|y_0|+1)},1 \bigg)$$
then $y\neq 0$ and
$$ \bigg| \frac{x}{y}-\frac{x_0}{y_0}\bigg|<\varepsilon.$$
Proof:
We note that
$$\bigg| \frac{x}{y}-\frac{x_0}{y_0}\bigg|=\bigg| x\cdot\frac{1}{y}-x_0\cdot\frac{1}{y_0}\bigg|$$
which is just $|xy-x_0y_0|$ of question 21 but with $y$, and $y_0$ replaced by $\frac{1}{y}$, and $\frac{1}{y_0}$ respectively, where $y\neq0$. Hence we require, by inequality (1) that
$$|x-x_0|<\min\bigg(\frac{\varepsilon}{2(1/|y_0|+1)}, 1 \bigg)\quad\text{and}\quad \bigg|\frac{1}{y}-\frac{1}{y_0}\bigg| < \frac{\varepsilon}{2(|x_0|+1)}. \tag{3}$$
Now in question 22, inequality (2) allows $\big| \frac{1}{y}-\frac{1}{y_0}\big|<\varepsilon$. However we must replace the $\varepsilon$ of inequality (2) with $\frac{\varepsilon}{2(|x_0|+1)}$ of (3) to give:
$$|y-y_0| <\min\bigg(\frac{|y_0|}{2}, \frac{\varepsilon |y_0|^2}{4(|x_0|+1)} \bigg).$$
To see why
$$|x-x_0|<\min\bigg(\frac{\varepsilon |y_0|}{4},1 \bigg)$$
is the wrong choice, we rework question 22:
\begin{align*}
\bigg| \frac{x}{y}-\frac{x_0}{y_0}\bigg|&= \frac{\left|xy_0-x_0y\right|}{\left|yy_0\right|}\\
&=\frac{ |x (y_0-y)- y(x_0-x)|}{|y||y_0|}\\
&\leq\frac{ | x ||y-y_0|}{|y||y_0|}+\frac{|x-x_0|}{|y_0|}\\
&\leq\frac{ (1+| x _0|)|y-y_0|}{|y||y_0|}+\frac{|x-x_0|}{|y_0|}\\
&<\frac{ (1+| x _0|)}{|y||y_0|}\cdot\frac{\varepsilon |y_0|^2}{4(|x_0|+1)}+\frac{|x-x_0|}{|y_0|}\tag{$\star$}\\
&<\frac{2}{|y_0|}\cdot\frac{ 1}{|y_0|}\cdot\frac{\varepsilon |y_0|^2}{4}+\frac{|x-x_0|}{|y_0|}\\
&=\frac{\varepsilon}{2}+\frac{|x-x_0|}{|y_0|}.
\end{align*}
Where we have used
$$\frac{1}{|y|}<\frac{2}{|y_0|}$$
from question 22 in the penultimate inequality. Now if $|x-x_0|<\min\big(\frac{\varepsilon |y_0|}{4},1 \big)$, we should have
$$\bigg| \frac{x}{y}-\frac{x_0}{y_0}\bigg|<\frac{\varepsilon}{2}+\frac{\varepsilon}{4}=\frac{3\varepsilon}{4}.$$
Now here could be a source of confusion because for sure $\frac{3\varepsilon}{4}<\varepsilon$, but that is not what is required. What is required is for $\big| \frac{x}{y}-\frac{x_0}{y_0}\big|<\varepsilon$, and as such we need to end up with two numbers $\varepsilon_1$ and $\varepsilon_2$, say, which add up to $\varepsilon$, which is some fixed number $>0$. The point is that when you assert something is less than $\varepsilon$ in the proof, you don't know if it is less than $\frac{3\varepsilon}{4}$, indeed, it could be greater than $\frac{3\varepsilon}{4}$, just as long as it is $<\varepsilon$.
The answer given in (3) works since
$$\frac{|x-x_{0}|}{|y_0|} <\frac{\varepsilon/|y_0|}{2(1/|y_0|+1)}< \frac{\varepsilon }{2},$$
which yields the desired $\varepsilon$ when plugged into $(\star)$.
Best Answer
The question has been broken into steps to try and help clarify what the essential ingredients in proving the result are, but you don't have to follow them if they don't seem intuitive. The result you want to prove is that, for any function $\phi$ which is non-increasing or non-decreasing on $[a,b]$, given any continuous function $f$, there is a $\xi \in (a,b)$ (depending on $f$) such that $$ \tag{$\dagger$} \int_a^b f(t)\phi(t)\mathrm{d}t = \phi(a).\int_a^{\xi} f(t)\phi(t)\mathrm{d}t + \phi(b).\int_{\xi}^b f(t)\phi(t)\mathrm{d}t $$
Part a) claims that knowing the result for nondecreasing $\phi$ implies the result for nonincreasing $\phi$, but it is also fine to try to prove it for one or the other, to begin with.
Similarly, the assumption that $\phi(b)=0$ makes some calculations easier, and therefore Spivak asks you to think about why the result for $\phi$ satisfying $\phi(b)=0$ implies the result for any nondecreasing $\phi$.
Thus parts (a) and (b) aim to reduce the general result to the special case where $\phi$ is nonincreasing and $\phi(b)=0$, and the condition that $\phi(b)=0$ means that the expression on the right-hand side of our desired identity becomes $$ \phi(a).\int_{a}^{\xi} f(t)\phi(t)\mathrm{d}t. $$ I think part of your confusion was the direction of the implication Spivak wants you to show -- if you know the existence of $\xi$ for any $\phi$ then it certainly gives you a suitable $\xi$ for a $\phi$ satisfying $\phi(b)=0$, but what you are asked to do is to show that, if you can prove there is a $\xi$ for which $\int_a^b f(t)\phi(t)\mathrm{d}t = \phi(a)\int_{a}^{\xi}f(t)\phi(t)\mathrm{d}t$ if $\phi$ is nonincreasing and $\phi(b)=0$, can you deduce the existence of $\xi$ satisfying $(\dagger)$ for an arbitrary nonincreasing $\phi$?
To give a hint for how to show part (b), for example, consider separately what happens to the lefthand side and right-hand side of $(\dagger)$ when you replace $\phi(t)$ with $\phi(t)+C$ for any $C\in \mathbb R$.