I am looking proof of above theorem, which avoids transfinite induction, and which is based on some tricks of Towers (for example, presented in Real and Complex Analysis by Rudin, or in book of Halmos on Set Theory).
The proof involves very different kind of arguments, which do not commonly appear in algebra, analysis or topology.
By looking at different references for "this proof", I was unable to understand the tricks used in proof.
Lemma: Suppose $\mathcal{F}$ is a nonempty collection of subsets of a set $X$ such that the union of every subchain of $\mathcal{F}$ belongs to $\mathcal{F}$. Suppose $g$ is a function which associates to each $A\in\mathcal{F}$ a set $g(A)\in\mathcal{F}$ such that $A \subset g(A)$ and $g(A)-A$ consists of at most one element. Then there exists an $A\in\mathcal{F}$ for which $g(A) = A$.
Proof of Lemma: Fix $A_0\in\mathcal{F}$. Call a subcollection $\mathcal{F}'$ of $\mathcal{F}$ a tower if it has the following three properties:
(a) $A_0\in\mathcal{F}'$.
(b) The union of every subchain of $\mathcal{F}'$ belongs to $\mathcal{F}'$.
(c) If $A\in \mathcal{F}'$, then also $g(A)\in \mathcal{F}'$.
The family of all towers is nonempty: if $\mathcal{F}_1$ is collection of members $A\in\mathcal{F}$ with $A_0\subset A$, then $\mathcal{F}_1$ is a tower.
Let $\mathcal{F}_0$ be the intersection of all towers. Then $\mathcal{F}_0$ is a tower, but no proper subcollection of $\mathcal{F}_0$ is a tower.
Claim: $\mathcal{F}_0$ is a chain (totally ordered subset): every element of $\mathcal{F}_0$ is comparable with every other element of $\mathcal{F}_0$.
Proof goes like this: observe that $A_0$ is comparable with all members of $\mathcal{F}_0$: because, $A_0$ is comparable with all members of $\mathcal{F}_1$ and $\mathcal{F}_0\subset\mathcal{F}_1$.
Let $\Gamma$ be a collection of comparable elements $C$ of $\mathcal{F}_0$: $\Gamma$ is non-empty because of previous remark.
For each $C\in\Gamma$, let $\Phi(C)$ be the collection of all $A\in\mathcal{F}_0$ such that either $A\subset C$ or $g(C)\subset A$. Then $\Phi(C)$ is a tower inside $\mathcal{F}_0$. [We skip proof]
By minimality of $\mathcal{F}_0$, we have $\Phi(C)=\mathcal{F}_0$ for every comparable element $C$ of $\mathcal{F}_0$.
Thus for every $C\in\Gamma$, every $A\in\mathcal{F}_0$ satisfies $A\subset C$ or $g(C)\subset A$. Hence $A\subset g(C)$ or $g(C)\subset A$. Which means, $g(C)$ is also comparable element of $\mathcal{F}_0$. Hence the set of all comparable elements of $\mathcal{F}_0$ is a tower; by minimality of $\mathcal{F}_0$, it should be whole $\mathcal{F}_0$.
Thus $\mathcal{F}_0$ is chain (i.e. totally ordered).
Let $A$ be the union of all members of $\mathcal{F}_0$. Then $A\in \mathcal{F}_0$ (by b) and $g(A)\in\mathcal{F}_0$ (by c). Hence $g(A)\subset A$ which is a subset of $g(A)$. Hence the proof.
Question 1. What is intuitive idea to bring towers in the arguments?
Question 2. The whole proof is running around the concept of different towers inside $\mathcal{F}$; but, I do not know whether this concept of tower is general or it is introduced only in this proof?