If the lattice $U$ satisfies the meet distributive law
$$x \wedge \bigvee_{i \in I} y_i = \bigvee_{i \in I} x \wedge y_i$$
where $(y_i)_{i \in I}$ is an arbitrary collection of elements of $U$, then "weak partitioning" implies "strong partitioning." More precisely, you only need the above to hold when the right hand side is $0$.

An example of a complete lattice where weak and strong partitioning are inequivalent is the lattice $U$ consisting of all closed subsets of $\{1,\frac12,\frac13,\ldots,0\}$ (as a subspace of $\mathbb{R}$) and the collection $S = \{\{\frac1n\}: n \geq 1\}$. The weak-partitioning property is easily verified since the points $\frac1n$ are isolated. The strong partitioning property fails for the two sets $A = \{\{\frac1{2n}\}: n \geq 1\}$ and $B = \{\{\frac1{2n+1}\} : n \geq 0\}$, for example, since $\bigvee A = \overline{\bigcup A}$ and $\bigvee B = \overline{\bigcup B}$ both contain the point $0$.

PS: In your formulation of weak and strong partitioning, I interpret $S$ as a collection of *nonzero elements* of $U$, since "nonempty subsets" doesn't make much sense in context.

of sets? (In the former case, changing your cups and caps with vees and wedges where appropriate would make the question easier to parse.) The answer to your question hinges on the validity of some distributive laws, these do hold in lattices of sets but not necessarily in general lattices. – François G. Dorais♦ Feb 18 '10 at 21:08