order theory - Weak partitioning vs. strong partitioning

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.