First of all, for what you write to make sense, you're not picking two partitions and defining a relation between those particular two partitions. You're defining a single relation on the set of all partitions.
Second, I think you're overcomplicating things by considering a partition to be an indexed family of subsets of $U$. Indeed, if you consider the assignment of indices to be part of the partition, then what you want to prove is not true, because then the two partitions
$ A_1 = \{1\}, A_2 = \{2,3,\ldots,n\} $ and $ B_1 = \{2,3,\ldots,n\}, B_2 = \{1\} $ would satisfy $A\succ B\succ A$, but $A\ne B$.
So we need to work with partitions being unordered collections of subsets of $U$, and your relation should then be defined as $ B\succ A \quad\iff \forall b\in B\; \exists a\in A: b\subseteq a$
In order to prove that this is antisymmetic, we assume $A\succ B\succ A$ and seek to prove that $A \subseteq B$. (Then, since $A$ and $B$ were arbitrary, and we also have $B\succ A\succ B$, the same argument shows $B\subseteq A$, so $A = B$).
In order to prove this, it is crucial that $A$ and $B$ are partitions, which requires among other things that (1) any two different members of $A$ must be disjoint, and (2) the empty set is not in $A$.
Now, we're assuming that $A\succ B\succ A$. To prove $A\subseteq B$ consider any $x\in A$. One of the $\succ$s gives us $y\in B$ such that $x\subseteq y$, and the other gives $z\in A$ sucht that $x\subseteq y\subseteq z$. But because $x$ and $z$ are both in $A$, they must be either equal or disjoint. Since $x$ is non-empty and $x\subseteq z$ they can't be disjoint, so $x=z$. But then $x\subseteq y \subseteq x$, and $y$ must equal $x$. Since $y$ was in $B$, we have proved $x\in B$.