Proposition 8.5.10. Let $X$ be a well-ordered set with an ordering relation $\leq$, and let $P(n)$ be a property pertaining to an element $n\in X$. (i.e., $\forall n\in X, P(n)$ is either a true or false statement) Suppose that $\forall n\in X$, if $\forall m\in X, P(m)$ is true s.t. $m<_{X}n$, then $P(n)$ is also true. Then $\forall n\in X, P(n)$ is true.
Proof: Let $X$ be a well-ordered set, then $\forall A\subseteq X$, $A$ has a minimal element; that is $\forall A\subseteq X\exists\alpha\in A\forall\alpha_1\in A(\alpha\leq_A\alpha_1)$. Suppose that no matter which $n\in X$ we choose, $\forall m\in X$ s.t. $m<_{X}n$, $P(m)$ is true will always imply $P(n)$ is true. Then if $n$ is not the maximal element in $X$, $P(n)$ would imply that $P(\overline{n})$ is also true, provided that $n<_{X}\overline{n}$. However, if there exist no maximal element in $X$, by the axiom of choice, we choose $\overline{n}$ to be the minimal element of some arbitrary $A_\overline{n}\subseteq X$. Since $\overline{n}$ is arbitrary then $\forall\overline{\eta}\in A_\overline{n}$, such that $\overline{n}<\overline{\eta}$, $P(\overline{\eta})$ is true. Since $A_\overline{n}$ is any subset of X then $\forall n\in X, P(n)$ is true.
It is not true that every subset of $X$ has a minimal element. This in spite of the fact that $X$ is well-ordered. Under that condition it is true however that every non-empty subset of $X$ has a minimal element.
You do not need the axiom of choice here.
Assume that some $n\in X$ exists for which $P(n)$ is not true. That means that the subset $A=\{n\in X\mid P(n)\text{ not true}\}$ is a non-empty subset of $X$ hence has a minimal element.
If $m$ denotes this element then $P(n)$ is true for all $n<m$ (if not then $m$ would not be minimal). But then we are allowed to conclude that also $P(m)$ is true, and a contradiction has been found.
We conclude that our assumption was wrong, so for every $n\in X$ the statement $P(n)$ is true.