Given a set of sets ${\cal A} = \{S_i\mid i\in {\cal B}\}$ and a binary relation $Con$ on $\bigcup {\cal A}$, a $Con$-choice on $\{S_i\mid {i\in F}\}, F\subseteq {\cal B},$ is a function $\epsilon\in \prod_{i\in F} S_i$ such that $\forall i,j\in {F}: i \not= j \rightarrow Con(\epsilon(i),\epsilon(j)).$
$CC({F},Con)$ denotes the set of all $Con$-choices on $\{S_i\mid{i\in F}\}$. ${\cal P}(X)$/${\cal P}^{fin}(X)$ denotes the set of all/all finite subsets of $X$.
Theorem 1 [Los, Ryll-Nardzewski, 1951] Let
$\bullet\ {\cal A} = \{S_i\mid i\in {\cal B}\}$ be a family of compact topological spaces, and
$\bullet\ Con$ be a symmetric relation on $\bigcup{\cal A}$ such that for each $i\not= j\in {\cal B}$, the set $CC(\{i,j\},Con)$ is closed in the product space $S_i\times S_j$.
If $\forall {F}\in {\cal P}^{fin}({\cal B}): CC({F},Con)\not=\emptyset$, then also $CC({\cal B},Con)\not=\emptyset$.
The following can hardly be true, but what is wrong with the proof? I would be very grateful for identifying the error.
For any set of sets ${\cal A} = \{S_i\mid i\in {\cal B}\}$ and a symmetric relation $Con$ on $\bigcup{\cal A}:$ if $\forall {F}\in {\cal P}^{fin}({\cal B}):CC({F},Con)\not=\emptyset$ then $CC({\cal B},Con)\not=\emptyset$.
Proof [?].
View each $S_i$ as the subset $\mathbf S^1_i = \{\{r\}\mid r\in S_i\}$ of the space $\mathbf{PS}_i = <{\cal P}(S_i),\tau_i>$, with Priestley topology $\tau_i$ given by the subbasis of clopens $\mathbf{[r]} = \{S\subseteq S_i\mid r\in S\}$ and $-{\mathbf{[r]}} = \{S\subseteq S_i\mid r\not\in S\}$, for each $r\in S_i$. $\mathbf{PS}_i$ is compact, and we consider ${\cal P\!A} = \{\mathbf{PS}_i\mid i\in {\cal B}\}$.
Defining ${PCon} \subseteq \bigcup{\cal P\!A} \times \bigcup{\cal P\!A}$ by $${PCon}(A_i,A_j) \Leftrightarrow A_i = \{p_i\}\in \mathbf S^1_i\land A_j = \{p_j\}\in \mathbf S^1_j \land Con(p_1, p_2), $$ makes ${PCon}$-choices and $Con$-choices correspond exactly to each other, i.e., $<\{p_1\},...,\{p_i\}> \in CC(F,{PCon})\ \Leftrightarrow\ <p_1,...,p_i>\in CC(F,Con)\}$.
For every pair ${i},{j}\in {\cal B}$, the set $$CC(\{i,j\},{PCon}) = \{<A_i,A_j> \in \mathbf{PS}_{i}\times \mathbf{PS}_{j} \mid {PCon}(A_i,A_j)\} \subseteq \mathbf S^1_i\times \mathbf S^1_j$$ is closed in the product space $\mathbf{PS}_{i}\times \mathbf{PS}_{j}$, because so is every $\mathbf Z\subseteq \mathbf S^1_i\times \mathbf S^1_j$. Namely, for every $x\not\in \mathbf Z$, there is an open $\mathbf O\ni x$ with $\mathbf O\cap \mathbf Z=\emptyset$. For any $<\{a\},\{b\}> \in (\mathbf S^1_i\times \mathbf S^1_j) \setminus Z$, the open $\mathbf O = \mathbf{[a]}\times\mathbf{[b]}$ gives $\mathbf O \cap (\mathbf S_i^1\times \mathbf S_j^1) = \{<\{a\},\{b\}>\}$, so that $\mathbf O \cap \mathbf Z = \emptyset$. For any other $<A,B> \not\in Z:|A|\geq 2$ or $|B|\geq 2$. Wlog., say $\{a_1,a_2\}\subseteq A$ and let $\mathbf B\subseteq \mathbf{PS}_j$ be any open with $B\in \mathbf B$. Then the open $\mathbf O = (\mathbf{[a_1]}\cap \mathbf{[a_2]}) \times \mathbf B$ contains $<A,B>$ and $\mathbf O \cap (\mathbf S_i^1\times \mathbf S_j^1) = \emptyset$, so that $\mathbf O\cap \mathbf Z = \emptyset$.
Since $\forall F\in {\cal P}^{fin}({\cal B}): CC(F,Con) \not=\emptyset$ so $CC(F,PCon) \not=\emptyset$ by (2). By Theorem 1, $CC({\cal B},{PCon})\not=\emptyset$ and hence $CC({\cal B},Con)\not=\emptyset$ by (2). $\Box$
I found a mistake in point 3, where the proof that Z is closed lacks one case, namely, of a point $<\emptyset,\{b\}> \not\in Z$, for which no required open exists.