My claim. Let $(X,<)$ be a well-ordered set, $Y$ and $<$-section and $f:X\rightarrow X$ an order-preserving function. Then
$$ Y'=\{x\in X : f(x)< x \} = \emptyset . $$
My proof.
1.- If $f=\mbox{id}_X$ then $Y'$ is celarly empty, since well-orders are asymmetric binary relations.
2.- If $f\not=\mbox{id}_X$ $Y'$ may be different from $\emptyset$. So suppose $Y'\neq \emptyset$. Then there exists a least element $y_0\in Y'$. However since it belongs to $Y'$, $f(y_0)<y_0$ (Does it shows $Y'$ is not a $<$-section?).
Moreover, since $f(y_0)\notin Y'$, then $f(y_0)<f(f(y_0))$. On the other hand, since $f$ is order-preserving, $f(f(y_0))<f(y_0)$.
Remark. The above statement is written in my book but it considers order-preserving mapps and it is used to show that if $Y$ is an $<$-section, then
$$\not\exists y\in Y:f(y)<y$$,
but I think this is a corollary. The theorem statement is much more strong, no?