property  R_\lambda A space X has property $R_\lambda$ if every family of its clopen sets of cardinality $\lambda$ has a subfamily of cardinality $\lambda$ which is either linked or disjoint. 
The following result was asserted by Bell [see A Ramsey theorem for polyadic spaces].
Theorem: Let $cf(\lambda) \geq \omega_1$ and let { $X_i : i \in I $} be a collection of Boolean spaces such that every finite product of them satisfies property $R_\lambda$. 
Then $\displaystyle{\prod_{i \in I} X_i}$ has property  $R_{\lambda}$.
He mentioned that its proof is an analogous to the proof of Noble–Ulmer theorem that shows a product is ccc iff every finite subproducts ccc. I tried  to proof it but i got a problem. Here is my try.
Proof: Suppose for a contradiction that there is a collection $\mathcal{O}=$ { $O_\alpha : \alpha < \lambda $ } of clopen sets in $\displaystyle{\prod_{i \in I} X_i}$ that does not contain any subcollection of cardinality $\lambda$ which is either linked or disjoint. So for each $\alpha$ there is a finite subset  $F_\alpha$ of $I$ and a nonempty basic clopen set $\displaystyle{\prod_{i \in I} U_{i}^{\alpha}} \subseteq O_\alpha$ such that $U_{i}^{\alpha} \neq X_i$ for all $i\in F_\alpha$ and $U_{i}^{\alpha} = X_i$ if $i\notin F_\alpha$. 
Let $\mathcal{G}=$ { $F_\alpha:~\alpha<\lambda$ }. This is a collection of finite sets, so it contains a $\Delta$-system of size $\lambda$. Say $\mathcal{D} \subseteq \mathcal{G}$ is a $\Delta$-system of size $\lambda$ with root $R$, and let $\mathcal{D}$ = { $ F_\alpha :\alpha \in A$ } for some set $A$ of size $\lambda$. Consider the subcollection { $O_\alpha : \alpha \in A$ } of $\mathcal{O}$. From the hypothesis it does not contain a subcollection of size $\lambda$ which is linked or disjoint.
Now consider $\{\displaystyle{\prod_{i \in R} {U_{i}^{\alpha}}} : \alpha \in A\}$ and claim it does not have a subcollection of size $\lambda$ which is linked or disjoint. Then we shall show that therefore the finite product $\displaystyle{\prod_{i \in R} X_i}$ violates property $R_{\lambda},$ which contradicts our hypothesis and hence this will complete the proof of the Theorem.
To prove the claim: Firstly, suppose that $B \subseteq A$ is of size $\lambda$ and $\{\displaystyle{ \prod_{i \in R} U_{i}^{\alpha}} :\alpha \in B\}$ is a linked subcollection of $\{\displaystyle{ \prod_{i \in R} U_{i}^{\alpha}}:\alpha \in A\}$. So for all $ \alpha \neq \beta \in B, ~~\mbox{there is an element }  z \in (\displaystyle{\prod_{i \in R} U_{i}^{\alpha} \cap \prod_{i \in R} U_{i}^{\beta}})$. Thus we can extend the range of $z$ to be an element of $\displaystyle{\prod_{i \in I} X_i}$ as follows:
For $i\in I\backslash R$ we have $U_{i}^{\alpha}=X_i$ or $ U_{i}^{\beta}=X_i$. Certainly $U_{i}^{\alpha} \cap U_{i}^{\beta} \neq \emptyset$. Let $u(i) \in U_{i}^{\alpha} \cap U_{i}^{\beta}$ and define $y \in \displaystyle{ \prod_{i \in I} X_{i}}$ by putting
$y(i)=z(i)$  if  $i\in R$  and $y(i)=u(i)$  otherwise. 
So, $y \in O_\alpha \cap O_\beta$ and hence the collection $\{ O_\alpha : \alpha \in B \}$ is a linked subcollection of $\displaystyle{\prod_{i \in I} X_i}$, contradicting the assumption that the family $\{O_\alpha : \alpha \in A\}$ does not have property $R_{\lambda}$.
Now for the second part, suppose that for some $A^* \in [A]^\lambda$ and for any $\alpha \neq \beta$ in $A^*$ we have $\displaystyle{\prod_{i \in R} U_{i}^{\alpha} \cap \prod_{i \in R} U_{i}^{\beta}}= \emptyset$. Since
 $$\displaystyle{\prod_{i \in R} U_i^\alpha \cap \prod_{i \in R} U_i^\beta = \emptyset \Longrightarrow \prod_{i \in I} U_i^\alpha \cap \prod_{i \in I} U_i^\beta}= \emptyset $$
the problem is how we get that $O_\alpha \cap O_\beta = \emptyset$ to get a contradiction with the fact that there is no disjoint subcollection of $O_{\alpha}\mbox{'s}$ of size $\lambda$. Any help please?
 A: You are not using the fact that your spaces are Boolean.  In your argument you pass from open set in the product to subsets in the usual basis of the product topology that have finite support.
But the original sets $O_\alpha$ are compact and therefore each $O_\alpha$ is the finite union of basic open sets.  This shows that each $O_\alpha$ depends only on finitely many coordinates to begin with.  I guess you can finish the proof from there.

Edit:  I was obviously not clear enough.  The argument should go as follows:  My claim is that all clopen subsets of the product of Boolean spaces only depend of finitely many coordinates.
This is because the clopen subsets of the product are compact and therefore they are finite 
unions of basic open sets.  So for each $\alpha$ there is a finite set $F_\alpha\subseteq I$
such that $O_\alpha$ only depends on the coordinates in $F_\alpha$.  
Now you go carry out the $\Delta$-system argument as before.  The projections of the $O_\alpha$'s to the product $\prod_{i\in R}X_i$ are clopen.  On this finite product you can thin out the collection of clopen sets so that you are left with $\lambda$ sets that are either linked or disjoint.  Now go back to the full space.  The corresponding subcollection of the $O_\alpha$ will have the desired property.
