Does double negation distribute over disjunction intuitionistically?

$\lnot \lnot (A \lor B) \to (\lnot \lnot A \lor \lnot \lnot B)$ is not intuitionistically acceptable. One way of seeing this is by considering the Heyting algebra whose elements are the open subsets of the unit interval $[0, 1] \subseteq \Bbb{R}$, with $A \lor B = A \cup B$, $A \to B = \mathsf{int}(A^c\cup B)$ and $\bot = \emptyset$ (see https://en.wikipedia.org/wiki/Heyting_algebra). In this Heyting algebra, $\lnot\lnot A$ is the interior of the closure of $A$ and $A \to B$ is $\top$ iff $A \subseteq B$. Hence if $A = [0, 1/2)$ and $B = (1/2, 1]$, $\lnot \lnot (A \lor B) = [0, 1]$ while $\lnot \lnot A \lor \lnot \lnot B = [0, 1] \mathop{\backslash} \{1/2\}$ and $\lnot \lnot (A \lor B) \to (\lnot \lnot A \lor \lnot \lnot B)$ is not $\top$.

By substituting ¬A for B in ¬¬(A∨B)→(¬¬A∨¬¬B) we can easily derive (¬¬A∨¬A) which is certainly not intuitionistically acceptable. Hence ¬¬(A∨B)→(¬¬A∨¬¬B) is also not acceptable.