Can someone please explain how it follows that $|N(S)|x \geq |S|x$? What I'm asking is why is it necessary to use the value of x to derive the inequality?
Theorem 5.2.7. Let G be a bipartite graph with vertex partition L, R where $|L| \leq |R|$. If G is degree-constrained, then there is a matching that covers L.
Proof.
The proof is by contradiction. Suppose that G is degree constrained but that there is no matching that covers L. By Theorem 5.2.5, this means that there must be a bottleneck $S \subseteq L$. Let x be a value such that $deg(l) \geq x \geq deg(r)$ for every $l \epsilon L$ and $r \epsilon R$. Since every edge incident to a node in S is incident to a node in N(S), we know that \begin{aligned} |N(S)|x \geq |S|x \end{aligned} and thus that \begin{aligned} |N(S)| \geq |S| \end{aligned} This means that S is not a bottleneck, which is a contradiction. Hence G has a matching that covers L.
EDIT:
For clarification (hopefully), this theorem builds on the definition of a bottleneck where:
\begin{aligned} G \ is \ the \ bipartite \ graph \ (V,E) \end{aligned}
So the graph is partitioned in to L and R, where:
\begin{aligned} S \subseteq L \end{aligned}
\begin{aligned} N(S) = \{r \ | \ \{s,r\} \ \epsilon \ E \ for \ some \ s \ \epsilon \ S \} \end{aligned}
Meaning that N(S) is a neighbor of some vertex in s, and:
\begin{aligned} |S| \geq |N(S)| \end{aligned}
implies that a bottleneck exists.
It's kind of subtle but everything is in 'Since every edge incident to a node in $S$ is incident to a node in $N(S)$'.
Here's how you can see this. Let $e(S)$ and $e(N(S))$ respectively denote the number of edges incident to some vertex of $S$ and $N(S)$.
Then, $e(N(S)) \geq e(S)$, since every edge counted in $e(S)$ has to be counted in $e(N(S))$ (as every edge with an endpoint in $S$ has an its other endpoint in $N(S)$.
We have $|S|x \leq e(S)$ since $x \leq deg(l)$ for all $l \in S$.
And we have $|N(S)|x \geq e(N(S))$ since $x \geq deg(r)$ for all $r \in N(S)$.
Putting all this together in the correct order, you get the inequality chain
$|S|x \leq e(S) \leq e(N(S)) \leq |N(S)|x$.