\noindent \textbf{Claim:} if $\varphi \vee \psi$ is an intuitionistic tautology, then so is at least one of $\varphi$ and $\psi$.
\begin{proof}
Suppose, for contradiction, that $\varphi \vee \psi$ is an intuitionistic tautology, but neither $\varphi$ nor $\psi$ is. Then, by definition, there exist Kripke models $(W_1,R_1,f_1)$, $(W_2,R_2,f_2)$ and states $w_1 \in W, w_2 \in W_2$ such that $(W_1,R_1,f_1),w_1 \nVdash \varphi$ and $(W_2,R_2,f_2),w_2 \nVdash \psi$. Now take the disjoint union of their isomorphic labeled copies. That is, consider $(W,R,f)$ with
\begin{itemize}
\item $W = W_1 \times {1} \cup W_2 \times {2}$,
\item $R = {\langle (w,1),(v,1) \rangle | (w,v) \in R_1 } \cup {\langle (w,2),(v,2) \rangle | (w,v) \in R_2 }$,
\item $f((w,i)) = f_i(w)$ for all $i \in {1,2}$ and $w \in W_i$.
\end{itemize}
\noindent Furthermore, we add a common root $u$ of $(w_1,1),(w_2,2)$ to the model, so we get $(W' = W \cup \{u\},R' = R \cup \{(u,u)\} \cup \{\langle u,(v_1,1) \rangle | R(w_1,1)(v_1,1)\} \cup \\ \{\langle u,(v_2,2) \rangle | R(w_2,2)(v_2,2)\}, f')$ with $f'(w) = f(w)$ for $w \neq u$, and $f'(u)= \emptyset$.
\end{proof}
\noindent It is straightforward to check that $(W',R',f')$ is still an intuitionistic Kripke model. $R$ is clearly still reflexive and transitive, and by adding $u$, we also added a reflexive arrow and all the transitive arrows to the successors of $(w_1,1)$ and $(w_2,2)$. Furthermore, monotonicity of $f$ still holds since $f(u) = \emptyset$. Thus, if $u R' (v,i)$, then we must have that $f(u) \subseteq f((v,i))$. Finally, since the successors of $w_1$, $w_2$ are unchanged, we have $(w,1) \nVdash \varphi$ and $(w,2) \nVdash \psi$.
\vspace{0.3cm}
\noindent Then it follows that $u \nVdash \varphi$ since $Ru(w_1,1)$ and $(w_1,1) \nVdash \varphi$ and $u \nVdash \psi$ since $Ru(w_2,2)$ and $(w_2,2) \nVdash \psi$. But then we have $u \nVdash \varphi \vee \psi$, which contradicts our assumption that $\varphi \vee \psi$ is an intuitionistic tautology.