Set that is not well-ordered I want to find a set that is transitive, every non-empty subset of it has $\in$-maximum and is not $\in$-well ordered.
Any hint?
 A: Let $S$ be a non-empty set such that every non-empty subset of it has an $\in$-maximum.
Then for $x,y \in S$ with $x \ne y$, we have either $x \in y$ or $y \in x$ because by hypothesis, $\{x,y\}$ has an $\in$-maximum. Therefore, $(S, \in)$ is a linear order. In the degenerate case that $S$ has only one element, clearly $(S, \in)$ is also a linear order.
Now suppose that a non-empty subset $T \subseteq S$ had no $\in$-smallest element. By the Axiom of Foundation, $T$ does have an $\in$-minimal element, i.e. an $x \in T$ with $x \cap T = \varnothing$.
But this then means that for all $y \in T$, we have $y \notin x$. Since $\in$ is a linear ordering on $S$, we conclude that $x \in y$ (if $x \ne y$). But then $x$ is an $\in$-smallest element; hence $S$ is well-ordered by $\in$.
Therefore, no set $S$ matching your conditions can exist. Note that the assumption that $S$ is transitive was not necessary to arrive at this conclusion.
A: I'm not an expert in Set Theory, but I'd say that if you're working in $ZF$, i.e, in particular, if you assume the Axiom of Foundation, then you can't find an example of what you're looking for. Indeed, if a set $x$ is transitive and each of its non empty subsets has $\in -$ maximum element, then trivially $x$ is totally (linearly) ordered by the membership relation. But in $ZF$ we can prove that a set $x$ is an ordinal (defined as a transitive set which is well-ordered by membership) if and only if it is transitive and totally ordered by $\in$ (you might try to show this as an exercise), and then there's nothing more to say.
A: $\{\varnothing,\{\varnothing\},\{\{\varnothing\}\}\}$
A: (I argue on the basis of $\sf{ZF}$).
Let $X$ be a set such that every nonempty subset has an $\in$-maximal element (or more precisely: greatest element).
Assume that $\in$ is not a wellorder on $X$.
As $a,b\in X$ implies that $\{a,b\}$ has an $\in$-maximal element, exactly one of $a\in b$, $a=b$, $a\ni b$ holds (the Axiom of Foundations forbids that two of these hold concurrently).
For $a,b,c\in X$ with $a\in b$ and $b\in c$, it is again the Axiom of Foundation that forbids $a=c$ and $c\in a$, hence $a\in c$. We conclude that $\in$ is a linear order on $X$.
Then the inverse relation $\ni$ is a well-order, hence there is an ordinal $\alpha$ and a bijection $f\colon \alpha\to X$ that inverts the order, i.e. $a\in b$ implies  $f(b)\in f(a)$.
Then a nonempty subset of $X$ without minimal element corresponds to a nonempty subset of $\alpha$ without maximal element. 
For such a subset to exist, we need $\alpha\ge\omega$.
The Axiom of Foundation demands that $S:=\{f(n)\mid n\in\omega\}$ has an element $f(n)$ with $f(n)\cap S=\emptyset$. As $f(n+1)\in f(n)\cap S$, we get a contradiction.
No such $X$ can exist (and we didn't even use transitivity).
