One can show inductively that $U(n+1)\subseteq U(n)$, and so the $T(n)$ are the differences in the descending hierarchy; thus, the question (T) amounts to whether the intersection of the $U(n)$ is empty. As you indicated in the comments, let's suppose that you use the usual Kuratowski encoding of ordered pair. In this case, both $x$ and $y$ are elements-of-elements of $\langle x,y,\rangle$. (And for most of the encodings of ordered pair, $x$ and $y$ are both in the transitive closure of $\langle x,y,\rangle$, which is the critical point.)
The answer to question 1 is Yes. If a set $a$ is in every $U(n)$, then we may unwrap $a$, since it is pair $a=\langle a_0,b_0\rangle$, and $a_0=\langle a_1,b_1\rangle$, and so on, with $a_n=\langle a_{n+1},b_{n+1}\rangle$, and $a_{n+1}\in\in a_n$, meaning that it is an element of an element, and this violates the well-foundedness of the $\in$ relation, contrary to the foundation axiom.
Similarly to your other recent question, the answer to question 2 is No. This is because it is relatively consistent with ZF- that there is a set $x$ such that $x=\{x\}$; such sets exist under the anti-foundation axiom. Note that $x=\{x\}=\{\{x\},\{x,x\}\}$$x=\{\{x\}\}=\{\{x\},\{x,x\}\}$, which is the same as $\langle x,x\rangle$. So $x$ is in every $U(n)$, violating (T).
I don't know the answer to question 3, but I expect that a solution will similarly as in your other question, which seems more fundamental to me.
