For a given set, does there always exists a well-ordering of the set of all its subsets which is stronger than the usual ordering (that is set-theoretic inclusion) of the sets of the subsets of the given set?
Again (now with symbols instead of words): Let $U$ be a set. Does there necessarily exists a well ordering of $\mathscr{P} U$ which is stronger than $\subseteq$ order?
I expect that there is a counter-example but haven't found one yet.
We can assume axiom of choice.
On
Consider the infinite descending chain of intervals $(-1/n,1+1/n)$ of real numbers. If $A\subseteq B\implies A\le B$, as in the comment, then this is an infinite descending chain in the supposed well-ordering, but has no smallest element. (If we wish, we can restrict to the rationals in the given intervals.)
No. For $n\in\Bbb N$ let $T_n=\{k\in\Bbb N:k\ge n\}$. Then
$$T_0\supsetneqq T_1\supsetneqq T_2\supsetneqq T_3\ldots$$
is an infinite descending chain in $\langle\wp(\Bbb N),\subseteq\rangle$ and would necessarily remain so in any extension of $\subseteq$. This will happen with any infinite set.