Similar to Is $ \forall x(P(x) \lor Q(x)) \vdash \forall x P(x) \lor \exists xQ(x) $ provable?, but with intuitionistic logic. I expect it is not, since I don't think the $\exists x Q(x)$ on the right is very "constructive", but curiously the finite version of this
$$((p_1\lor q_1)\land(p_2\lor q_2))\to((p_1\land p_2)\lor(q_1\lor q_2))$$
is intutitionistically provable, since you can break up the left side into four cases and verify that each satisfies the right side.
On
Just for the record, here's an answer using the open set model. What we have to do is find families of open sets $P_i$ and $Q_i$ in some topological space $X$ such that: $$ Int(\bigcap_i(P_i \cup Q_i)) \not\subseteq Int(\bigcap_i P_i) \cup (\bigcup_i Q_i) \tag*{(*)} $$ To do this, take $X$ to be the unit interval $[0, 1]$ and for $i = 1, 2, \ldots$ define $$ P_i = [0, \frac{1}{i}) \\ Q_i = (\frac{1}{i+1}, 1] $$ Since $X = P_i \cup Q_i$ for any $i$, the left-hand side of $\mbox{(*)}$ is equal to $X$. However, $\bigcap_i P_i = \{0\}$, which has empty interior, so the right-hand side of (*) is $\bigcup_i Q_i = (0, 1] \not \supseteq X$.
No, it is not intuitionistically valid. Here is a Kripke model where it doesn't hold:
There are two frames, $w$ and $u$ with $w<u$.
In $w$ the universe is $\{\mathtt a\}$ and $P(\mathtt a)$ is true.
In $u$ the universe is $\{\mathtt a,\mathtt b\}$, and $P(\mathtt a)$ and $Q(\mathtt b)$ are true.
Then both frames satisfy $\forall x(P(x)\lor Q(x))$.
However $w$ doesn't satisfy $\forall x\,P(x) \lor \exists x\,Q(x)$. The first disjunct is false because the intuitionistic interpretation of $\forall$ must be true in all reachable frames (and it clearly isn't true in $u$), and the second is clearly false in $w$.