I have never seen this claim in any textbook; in any case it's wrong. The claim seems to be that if you have two fractions $\frac{a}{b}$ and $\frac{c}{d}$ with $a < b$ and $c < d$, then $|a - b| < |c - d| \iff \frac{a}{b} < \frac{c}{d}$.
This is false. We can write $\frac{a}{b} = 1 - \frac{b-a}{b}$ and $\frac{c}{d} = 1 - \frac{d-c}{d}$, so you're comparing $\frac{b-a}{b}$ and $\frac{d-c}{d}$ (whichever is greater, the corresponding fraction is smaller). The first numerator may be smaller than the second, but the actual comparison of these fractions can of course go either way.
For example,
Here is one with $b - a < d - c$ but $\frac{a}{b} > \frac{c}{d}$: consider $\frac{2}{3} > \frac{3}{5}$.
Here is one with $b - a < d - c$ but $\frac{a}{b} = \frac{c}{d}$: consider $\frac{2}{3} = \frac{4}{6}$.
Here is one with $b - a < d - c$ but $\frac{a}{b} < \frac{c}{d}$: consider $\frac{2}{3} < \frac{5}{7}$.
So all results are possible; the test is nonsense.
Edit: Just for fun/completeness, here is a table showing pairs $(\frac{a}{b}, \frac{c}{d})$ with each possible combination of the two comparisions:
$$\begin{array}{c|c|c|c}
& \frac{a}{b}<\frac{c}{d} & \frac{a}{b}=\frac{c}{d} & \frac{a}{b}>\frac{c}{d}\\
\hline \\
b-a<d-c & \frac23,\frac57 & \frac23,\frac46 & \frac23,\frac35\\
\hline \\
b-a=d-c & \frac23,\frac34 & \frac23,\frac23 & \frac23,\frac12\\
\hline \\
b-a>d-c & \frac57,\frac23 & \frac46,\frac23 & \frac35,\frac23 \\
\end{array}$$
(If you want examples involving fractions greater than $1$, turn each of the fractions upside down. Each of the inequalities between the fractions will reverse direction, so you'll still have a complete set of examples.)
Edit: On looking at that segment of the video, it's possible (not very clear) that what he may have been saying is equivalent to the following claim, which is true: if you have two fractions $\frac{a}{b}$ and $\frac{c}{d}$ with $a > b$ and $c > d$, and if $a - b < c - d$ and $b > d$, then $\frac{a}{b} < \frac{c}{d}$. Proof:
$$\frac{a}{b} = 1 + \frac{a-b}{b} < 1 + \frac{c-d}{b} < 1 + \frac{c-d}{d} = \frac{c}{d}$$
With fractions less than $1$, the corresponding statement would be that if you have two fractions $\frac{a}{b}$ and $\frac{c}{d}$ with $a < b$ and $c < d$, and if $b - a < d - c$ and $b > d$, then $\frac{a}{b} > \frac{c}{d}$:
$$\frac{a}{b} = 1 - \frac{b-a}{b} > 1 - \frac{b-a}{d} > 1 - \frac{d-c}{d} = \frac{c}{d}$$
But these are so many conditions on the hypothesis that I wonder how often it will be useful.
Best Answer
$$\left|\frac pq-\frac{22}7\right|=\left|\frac{7p-22q}{7q}\right|$$
The point is to solve the Diophantine equation $$7p-22q=\pm1$$ and get the greatest possible value for $q$.
It turns out that $7\cdot 311-22\cdot99=-1$, so your fraction is $\frac{311}{99}$