Proof of the disjunction property I am trying to prove the disjunction property "if $\,\vdash\phi\lor\psi$ then $\,\vdash\phi$ or $\,\vdash\psi$" for intuitionistic propositional logic.
So far I thought about choosing two non-tautologies $\phi$ and $\psi$ with two Kripke countermodels:
$$(W, R_W, f_W)~~\mbox{such that}~~\exists (w\in W)(w\not\Vdash\phi)$$ 
$$(V, R_V, f_V)~~\mbox{such that}~~\exists (v\in V)(v\not\Vdash\psi)$$ 
Then create a new Kripke model $(W\cup V\cup\{u\}, R, f)$ with $R(u,w)$ and $R(u,v)$. 
My reasoning is that $w\not\Vdash\phi\implies u\not\Vdash\phi$ and $v\not\Vdash\psi\implies u\not\Vdash\psi$, and therefore $u\not\Vdash\phi\lor\psi$, which provides a countermodel for $\phi\lor\psi$ being a tautology and $\phi$ and $\psi$ not being tautologies.
Is this sufficient to be a proof? For me it's unclear whether adding a world $u$ changes which sentences $w$ forces. So if I add a world $u$, can I be certain that still $w\not\Vdash\phi$?

Yes, your proof works; you have only to specify that the two countermodels must have disjoint frames $\langle W_1, R_1 \rangle$ and $\langle W_2, R_2 \rangle$. 
The new model will have $W = \{ w_0 \} ∪ W_1 ∪ W_2$ and the "extended" accessibility relation will be : 

$xRy$ iff $x=w_0$ or $xR_1y$ or $xR_2y$.

