Why in the $n$-th Farey sequence the product of the denominators of $2$ adjacent fractions exceed $n$ ($0$ and $1$ are excluded) ?
I have a theorem of Hurwitz which states:
For every irrational number $\alpha\in\mathbb R$ there exists infinitely many rationals $p/q$, such that
$\lvert\alpha-\frac pq\rvert<\frac{1}{\sqrt{5}q^2}$
and here, in the proof of Corollary $1.6$, if $p/q<\alpha<p'/q'$, then $p/q$ and $p'/q'$ are not in $S$ ?
I know that, if $p/q$ and $p'/q'$ are adjacent then; $\lvert\frac pq-\frac {p'}{q'}\rvert=\frac{1}{qq'}$ so according to the proof $qq'>n$, but why ?
You are likely misunderstanding what the author writes in the proof of Hurwitz's Theorem.
Specifically, he says, "if $S$ a finite set of rational numbers, then we can take $n$ large enough...", etc.
This means that for the given set of rationals $S$ (not necessarily a set of Farey fractions) he is considering the Farey sequence with $n$ large enough, so that it holds: $|\gamma-\alpha|>\frac{1}{n}$, for every $\gamma\in S$. In other words, by construction he picks suitable $n$, so that the gaps between the terms of the Farey sequence he considers are at most $1/n$.
That's the same as saying, given a finite set of rationals $S$, you can find a Farey sequence which partitions your total interval finer than how the set $S$ does it.
But then, by that very consideration, the gaps between any two successive Farey numbers in his sequence, will be:
$$\left|\frac{p}{q}-\frac{p'}{q'}\right|=\frac{1}{qq'}\lt \frac{1}{n}\lt |\gamma-\alpha|$$
which means of course,
$$qq'>n$$