I am having difficulties in understanding this proof of the Axiom of choice implying Tukey's lemma presented in the book. Real and Abstract Analysis by Hewitt and Stromberg:
The method of proof seems to be 'by Contradiction'. I am having difficulties in spotting where and for what did we achieve a contradiction when assuming there exists such a nonvoid family of finite character having no maximal member. How did the idea of spotting a f-inductive set come about and what could be a general and if possible shorthand and a bit elementary synopsis of this proof.
I have taken a look at the post tukeys lemma by axiom of choice. And the answers used ideas of transfinite induction and ordinals that have not yet been introduced in the text.
I would greatly appreciate any help or suggestions regarding this.
A subset of $\mathscr F$ being $f$-inductive means it's closed under applications of $f$ and under taking chain unions (as well as containing the empty set). Then you can think of the smallest $f$-inductive set $\mathscr I_0$ as being the set generated by $\{\emptyset\}$ under these operations, much as one can talk about the $\sigma$-algebra generated by a given collection (here the operations are countable unions and complements) or a subgroup generated by some elements (where the operations are the group multiplication and inverse).
You ask in a comment where the finite character property is used, and the crucial place is in showing that $\mathscr F$ is $f$-inductive. Note that until this it shown, it's not clear that there are any $f$-inductive sets at all. (But from there, the fact that there is a smallest one is then easy, since closedness under operations is clearly preserved when you take intersections.) Namely, it is crucial to the proof that $\mathscr F$ is closed under chain unions: any finite subset $B\subseteq \bigcup_i C_i$ is a finite subset of one of the $C_i$, thus (by finite character) $B\in \mathscr F$, and then (again by finite character), $\bigcup_i C_i\in \mathscr F.$
Now, think of what $\mathscr I_0$ looks like. The general way you obtain closure under some operations is you just iterate them a bunch until it's closed. So the first thing we might do is something like $$\emptyset\subsetneq f(\emptyset)\subsetneq f(f(\emptyset))\subsetneq\ldots$$... all of these definitely are in $\mathscr I_0$. But then, note that $\{f^{(n)}(\emptyset): n\in \mathbb N\}$ is a chain, so by the previous paragraph, $\bigcup_n f^{(n)}(\emptyset)\in \mathscr F$. Call this $f^{(\omega)}(\emptyset),$ and we have $f^{(\omega)}(\emptyset)\in \mathscr I_0$ and $f^{(n)}(\emptyset) \subsetneq f^{(\omega)}(\emptyset)$ for all $n$.
Then we can keep iterating applying $f$: $$f^{(\omega+1)}(\emptyset) := f(f^{(\omega)}(\emptyset))\supsetneq f^{(\omega)}(\emptyset)\\f^{(\omega+2)}(\emptyset) := f(f^{(\omega+1)}(\emptyset))\supsetneq f^{(\omega+1)}(\emptyset)$$ and so on and then once we're done with that, we have an even bigger chain, and we can take the union of that and keep going.
In fact - due to the crucial property that $\mathscr F$ is closed under chain unions due to its finite character and that it's closed under $f$ by definition/assumption- this process is never going to stop. This is the idea of the transfinite recursion proof you mentioned. Transfinite recursion formalizes 'iterating this process forever', and then we invoke Hartogs' theorem (which implies we can't iterate forever and stay confined to a set) to get a contradiction.
Alas, the proof in the book isn't by transfinite recursion (or more accurately, as Asaf alludes to in the comments, it is a proof by transfinite that is stubbornly refusing to be one), but looking at it this light does help make sense of the comparatively cryptic argument given. Notice that at each stage in building the closure of $\{\emptyset\},$ we simply made a longer chain and there was never any reason closure forced us to branch off into incomparable sets. That's pretty crucial: $\mathscr I_0$ should be a chain!
But then we get a contradiction. If $\mathscr I_0$ were a chain, then by closure, its union would need to be in $\mathscr I_0$ as well, but its union is a strict superset of any element of $\mathscr I_0$, so it can't be in $\mathscr I_0$.
And you can see at the bottom that indeed the apex of the proof is that $\mathscr I_0$ is a chain and that it generates a contradiction in this manner... and the elaborate couple of paragraphs before are all to show this without resorting to transfinite recursion.
Moreover, with the picture above in mind, it should be much more clear when you look at the definitions that you should expect $\mathscr H=\mathcal I_0$ and $\mathscr G_A=\mathcal I_0$ for any $A\in \mathscr H$. For example, for each new set $A:=f^{(\alpha)}(\emptyset)$ we add to our chain we have $f(B)\subseteq A$ for any previously added set, since we add new elements by applying $f$. The work still needs to be done to prove these things (or alternatively to develop the theory of transfinite recursion and prove it that way instead), but that should give you some intuitive background.
This is already long enough, but as a side note, it's good to step back and think about what Tukey's lemma does for us as a version of choice and how it fits in with the others.
A lot of books use Zorn's lemma predominantly, which says that any nonempty poset where each chain has an upper bound has a maximal element. But many of the key landmark proofs use it in a remarkably similar way (every vector space has a basis, every ring has a maximal ideal, every consistent set of sentences has a maximal consistent extension, to name a few).
For instance for vector spaces having bases you partially order the linearly independent sets by $\subseteq$, then show that it is closed under chain unions as follows: If the chain union $\bigcup_i C_i$ were linearly dependent, then there would be a finite number of vectors in $\bigcup_i C_i$ that linearly combine to zero, but then those finitely many vectors would need to all be in some $C_i$ in the chain, contradicting its linear independence. Sound familiar? What we're using is exactly that the collection of linearly independent subsets has finite character.
The argument for maximal ideals in rings is the exact same, just replace 'linearly independent' and 'there would be a finite number of vectors in $\bigcup_i C_i$ that linearly combine to zero' by 'generating a proper ideal' and 'there would be a finite number of ring elements in $\bigcup_i C_i$ that linearly combine to $1$'. And for a maximal consistent set of sentences replace by "consistent", and "any inconsistency proof uses only a finite number of sentences in $\bigcup_i C_i$". These of course are just manifestations of finite character as well.
So Tukey's lemma is just factoring this ubiquitous pattern repeated in many many proofs out of the proofs and into Zorn's lemma.
(For this reason I would prefer an approach that proves Zorn first, then Tukey from it if the intention is to exploit the ease of use of Tukey for typical AC applications in algebra... I would hope that order of things would make the lightbulbs light up for people who have only seen Zorn's lemma.)