Let $X$ be a set. Recall that an $U\subseteq {\cal P}(X)$ is upwards-closed if $A\in U, A\subseteq B \Rightarrow B\in U$ for any $A,B \subseteq X$. I'll write "ucs" as a shorthand for upwards-closed set.
Clearly, an ucs on $X$ contains $\emptyset$ iff it coincides with the whole of ${\cal P}(X)$ ; in this case, or if it is empty, I call it degenerate.
For any $Y \subseteq {\cal P}(X)$, define $${\cal T}(Y)=\lbrace Z \subseteq X \ | \ \forall y\in Y, Z \cap y \neq \emptyset\rbrace \tag{1}$$
(the so-called "transversals" on $Y$). This is clearly an ucs on $X$, and it is nondegenerate as soon as $\emptyset\not\in Y$ and $Y\neq \emptyset$.
My question : Can one show in ZFC that ${\cal T}({\cal T}(U))=U$ for any nondegenerate ucs $U$ (or find a counterexample) ? Can one show it in ZF ?
My thoughts : There is a dual construction to (1) : for $F\subseteq {\cal P}(X)$, let $${\cal U}_X(F)=\lbrace Z \subseteq X \ | \ \exists f\in F, f \subseteq Z\rbrace\tag{2}$$ This is clearly an ucs on $X$.
I say that $U$ is "finitely generated" when it is of the form ${\cal U}_X(F)$ for a finite $F\subseteq {\cal P}(X)$. In that case, the study of ${\cal T}(U)$ reduces to the computation of transversals over a finite set. Indeed, write $F=\lbrace F_1,F_2,\ldots,F_r \rbrace$ where $F_i \subseteq X$. Then there are disjoint subsets $G_1,G_2,\ldots,G_s$ (with $s\leq 2^r$) of $X$ such that each $F_i$ is a finite union of $G_j$'s', say $F_i=\bigcup_{j\in H_i} G_j$ where $H_i \subseteq \lbrace 1,2,\ldots, s \rbrace$. Let $X'=\lbrace 1,2,\ldots, s \rbrace$ and let $I'$ be the ideal on $X'$ defined by $U'={\cal U}_{X'}(\lbrace H_1,H_2,\ldots,H_r\rbrace)$. Then, it is easy to check that
$$ {\cal T}(U)={\cal U}(L) $$
where
$$ L=\bigg\lbrace \bigcup_{j\in U} G_j \bigg| U \in I' \bigg\rbrace $$
So if the result is true when $X$ is finite, it will automatically follow for finitely generated upward-closed sets.
Update 09/08/2019 : A concrete example : if $X=\lbrace 1,2,3,4,5,6 \rbrace $ and $$ U={\cal U}_X(\lbrace \lbrace 1,2,3 \rbrace, \lbrace 3,4,5 \rbrace, \lbrace 1,5,6 \rbrace \rbrace)$$
then
$$ {\cal T}(U)={\cal U}_X(\lbrace \lbrace 1,3 \rbrace, \lbrace 1,4 \rbrace, \lbrace 1,5 \rbrace, \lbrace 2,5 \rbrace, \lbrace 3,5 \rbrace, \lbrace 3,6 \rbrace, \lbrace 2,4,6 \rbrace \rbrace)$$
In any similar example where $X$ is finite, the identity ${\cal T}({\cal T}(U))=U$ is readily verified, but I was unable to construct a general proof.
Yes, I even was an opponent of a PhD thesis"Algebraic and topological structures on the superextensions" by Volodymyr Gavrylkiv (English abstract starts at p. 17). Proposition 3.6.1 of the thesis is stated for topological spaces and directly implies the positive answer for a finite set $X$, considered as a topological space endowed with a discrete topology. But its proof also can be simplified to obtain a positive answer for the case of (not necessarily finite) sets as follows.
Let $U$ be a non-degenerate ucs of subsets of a set $X$. Clearly, $U\subset {\cal T}({\cal T}(U))$. On the other hand, assume that there exists a set $F\in {\cal T}({\cal T}(U))\setminus U$. Since $F\not\in U$, $X\setminus F\in {\cal T}(U)$. But $F\in {\cal T}({\cal T}(U))$, so $F\cap (X\setminus F)\ne\varnothing$, a contradiction.