Let $W$ be well-ordered and $f$ increasing. If $z$ is the smallest $x$ such that $f(x)<x$, why does $f(f(z))<f(z)$?

Question

I am currently working on Set Theory of Thomas Jech, and I have a question about Lemma 2.4 which is :

Lemma
If $(W, <)$ is a well-ordered set and $f:W \rightarrow W$ is an increasing function, then $f(x)\geq x$ for each $x \in W$.

Proof
Assume that the set $X=\{x \in W : f(x) < x\}$ is nonempty and let $z$ be the least element of $X$. If $w=f(z)$, then $f(w)<w$, a contradiction.

I have a problem understanding the proof. I've found that the proof makes sense if $w \in X$ since
(i) $z\in X \rightarrow f(z)<z$ and
(ii) $z$ is least element of $X \rightarrow (\forall x \in X)z \leq x)$
(iii) $f$ is increasing $\rightarrow (\forall x \in X)f(z) \leq f(x) $
(iv) $w \in X \rightarrow f(w) < w$
But by (iii) and $w \in X$, $f(z) \leq f(w) \iff w \leq f(w)$. This contradicts with (iv).

However, I can't figure out why $w \in X$. Why is it? Or is there any mistake with my proof?

Answer 1

Since $f$ is increasing and $w<z$, $f(w)<f(z)=w$.