Let $B$ be a Boolean algebra and $S \subseteq B$ be a subset that generates $B$. Is it the case that every filter $x$ of $B$ is equal to the filter generated by $x \cap S$? What if $S$ itself is a lattice with respect to the induced partial order?
Of course, every element of $x$ is $t(\bar a)$ where $\bar a \in S$ and $t(\bar x)$ is a term in the signature of Boolean algebra, and $t$ can be put in terms of nice normal forms, but I can't apply those facts to this problem.
Let $B$ be the Boolean algebra of finite/cofinite subsets of $\mathbb N$.
Then $B$ is generated by the set $S$ of its atoms, since each set is a union of singletons.
Now take $x$ to be the filter of cofinite subsets of $\mathbb N$; it follows that $x \cap S = \varnothing$, so $x$ is not generated by that intersection.
The same argument applies if you take $S$ to be the lattice of finite subsets.