I read a proof that went along these lines:
"Let $(X, \le)$ be a woset. Let $E$ be a subset of $X$ such that:
- the minimal element of $X$ is a member of $E$
- for any $x \in X$, if $\forall_y[y \lt x \rightarrow y \in E]$, then $x \in E$
Then, $E = X.$ Proof:
Suppose that $E \ne X$. Then, let $x$ be the minimal element of $X - E$. Since, it is an element of $X - E$, it can not be an element of $E$. But, any $y \lt x$ would imply $y \in E$, because $x$ is the minimal element of $X - E$, so any element $y$ such that, $y \lt x$ would have to be in $E$. If it wasn't, $x$ wouldn't be the minimal element of $X -E$. Since $y \lt x$ implies $y \in E$, then by the second condition, $x \in E$ and a contradiction arises, so $X = E$."
But, then the claim was made that this being true allows proof by induction on any well founded set. I understand the proof, but I don't see how this theorem and induction are related. Could someone explain?