Relation between a fixed point and being a well-order I've been trying to prove the following, but with no particular success:

Given a linear order $\leq$ on $A$, define $\pi:2^A\to 2^A$ by $X\mapsto\{y\in A: (\forall x < y)(x\in X) \}$. Let $A_0$ be the least fixed point of $\pi$. Prove that $x\in A_0$ iff $\{(a,b)\in A\times A: a\leq b < x\}$ is a well order.

For example, suppose that $\{(a,b)\in A\times A: a\leq b < y\}$ is a well order. I must prove that $y\in A_0$.
Here's my thought process: what does it mean that $y\in A_0$? $A_0$ is a fixed point for $\pi$ iff $A_0=\{y\in A:(\forall x < y)(x\in A_0)\}$. So the condition that $y$ lies in a fixed point $A_0$ means that $(\forall x < y)(x\in A_0)$. But to chech this condition, I'd need to check the sub-condition $x\in A_0$, which again means that $(\forall t < x)(t\in A_0)$, and now the same problem arises with checking the sub-condition $t\in A_0$...
Even setting this aside, I must use somehow that $\{(a,b)\in A\times A: a\leq b < y\}$ is a well order. So I'd think I need to guess some particular non-empty subset of $A$ that I need to consider, and then I need to take a least element in that subset. I don't really see which subset to consider, and how the "leastness" property would help.
For the other direction, I have the same problem as described at the beginning. When trying to use the condition that $y\in A_0$, an infinite chain of conditions arises...
 A: For the reverse direction, since you have a well-ordering, you can use induction to turn your "infinite descending" argument into a proof that works.  That is, let $B=\{a\in A: a<y\}$ and then prove that $a\in A_0$ for all $a\in B$ by induction on $A$ (since you know $B$ is well-ordered by $\leq$).  In this proof by induction, you take $b\in B$ and get to assume every $a$ such that $a<b$ is in $A_0$, and want to prove $b\in A_0$ as well.  But this is immediate since the induction hypothesis exactly tells you that $b\in \pi(A_0)=A_0$.
(Notice that in this direction you don't use the leastness of $A_0$, only that it is a fixed point.  This shouldn't be surprising: since $A_0$ is the least fixed point, showing $y\in A_0$ is equivalent to showing that $y$ is in every fixed point of $\pi$, so all you should expect to use about $A_0$ in this proof is that it is a fixed point.)
For the forward direction, you'll need a different approach, since this time you do need to use the fact that $A_0$ is the least fixed point of $\pi$, rather than just that it is a fixed point.  (For instance, if you only knew it was a fixed point, then it could just be all of $A$, and $\leq$ certainly doesn't have to be a well order on all of $A$.)  So, how do you use an assumption like this?  Well, since it is the least fixed point, it is contained in any other fixed point.  So to show $A_0$ is contained in some other set, it's enough to just show that set is a fixed point.  That is, if $A_1$ is the set of all $x$ such that $\{(a,b)\in A\times A: a\leq b < x\}$ is a well order, if you can show that $\pi(A_1)=A_1$, then you can conclude that $A_0\subseteq A_1$.
