If for any partitions $A$ and $B$ we write $A\preceq B\iff \forall Y\in B\exists X\in A:X\subseteq Y$ then does it follow from Zorns lemma that for any set of partitions $S$ the preorder $P=(S,\preceq )$ has a minimal element?
My reasoning is that for any chain $C$ of $P$ if we define the relation $R=\small\bigcap_{P\in C}\left(\bigcup_{Q\in P}Q\times Q\right)\subseteq S\times S$ then there exists a unique partition $P'\in C$ such that $R=\bigcup_{Q\in P'}Q\times Q$ thus $P'$ is a lower bound of this chain therefore this construction shows every chain in $P$ has a lower bound which by Zorn's lemma means $P$ must have a minimal element. Or have I made an error here?
No. The partition $P'$ you describe exists, but there is no reason to think that $P'\in S$. Indeed, if you take a descending sequence of partitions $A_0\succ A_1\succ A_2\succ\dots$ and take $S=\{A_n\}_{n\in\mathbb{N}}$ then $S$ itself is a chain which has no lower bound in $S$, and $S$ has no minimal element. For an explicit example of a partition, you can start with the 1-set partition of an infinite set and then break off singletons from it one at a time.
Or, for a simpler example, if $S=\emptyset$ then the empty chain has no lower bound in $S$, and again $S$ has no minimal element.