The infinite distributive law on a join-complete lattice $L$ is as follows:
$\displaystyle a \wedge\left( \bigvee_{b \in B} b \right) = \bigvee_{b \in B}(a \wedge b) $ for all $a \in L$ and $B \subset L$
Note any collection of sets satisfies the infinite distributive law if the meet and join operations are intersection and union respectively. Hence any concrete lattice $-$ i.e. a lattice of sets taken under union and intersection $-$ satisfies the infinite distributive law if it is closed under arbitrary unions.
In Peter T. Johnstone's Stone Spaces, Exercise 2.5 is to prove every distributive lattice $L$ is isomorphic to a concrete lattice $\Phi$. In this case we identify each $x \in L$ with the set of all morphisms $\phi \colon L \to \{0,1 \}$ for which $\phi (x) = 1$. Some form of Choice is needed to guarantee the existence of sufficiently many morphisms.
If we assume $L$ is finitely distributive and join-complete $-$ has arbitrary suprema $-$ then $\Phi$ will be closed under arbitrary unions. But then $\Phi$ hence $L$ satisfies the infinite distributive law. The conclusion is that join-complete lattices satisfy the infinite distributive law.
Is there a flaw to this reasoning? I cannot see one, but the conclusion is not what the book has led me to expect.
Let $L$ be the lattice of closed subsets of $\Bbb R$. Finite meets and joins in $L$ are just unions and intersections, so $L$ is distributive, and for arbitrary $F\subseteq L$ we have $\bigvee F=\operatorname{cl}\bigcup F$, so $L$ is join-complete. However,
$$\{0\}\land\bigvee_{n\in\omega}\left[2^{-n},1\right]=\{0\}\ne\varnothing=\bigvee_{n\in\omega}\big(\{0\}\land\left[2^{-n},1\right]\big)\;.$$