Interweaving two indexed families of filters Conjecture
  Let $U$ be an (infinite) set. Let $f$ be an $N$-ary (where $N$ is an
  arbitrary index set) relation on $U$ (that is a set of functions $N
  \rightarrow U$).
Let $\mathcal{L}_0$, $\mathcal{L}_1$ be $N$-indexed families of filters (on
  $U$).
I denote $\operatorname{up} \mathcal{L} = \prod_{i \in N}
  \mathcal{L} (i)$.
Let $\forall X \in \operatorname{up} \mathcal{L}_0 \cap \operatorname{up} \mathcal{L}_1 :
  \left( \prod_{i \in N} X_i \right) \cap f \neq \emptyset$.
Prove (or disprove) that
  $$ \exists c \in \{ 0, 1 \}^N \,
\forall Y\in \prod_{i\in N}\mathcal{L}_{c (i)} (i) :
\left( \prod_{i \in N} Y_i \right) \cap f \neq \emptyset . $$
 A: Something's wrong in the last formula of the question.  The elements of $f$ are functions  $N\to U$, but the elements of $\prod_{i\in N}(\mathcal L_{c(i)}(i))$ are functions on $N$ whose values are subsets of $U$, elements of the filters $\mathcal L_{c(i)}(i)$, not elements of $U$.  I'm going to assume that you meant that for every choice of sets $Y_i\in\mathcal L_{c(i)}(i)$ (for all $i\in N$), $(\prod_{i\in N}Y_i)\cap f\neq\emptyset$.
If that's the intended question, then here's a counterexample.  Let $U$ and $N$ both be the set of natural numbers.  Let $\mathcal L_0(0)=\mathcal L_1(0)=$ the filter of cofinite subsets of $U$.  For all $n>0$, let $\mathcal L_0(n)$ be the principal filter generated by $\{0\}$ and let $\mathcal L_1(n)$ be the principal filter generated by $\{1\}$.  Let $f$ be the family of those functions $x:N\to U$ such that $x(n)=0$ whenever $1\leq n\leq x(0)$, and $x(n)=1$ whenever $n>x(0)$.  (The only important thing here is that $x(0)$ determines the values of $x$ at all points other than 0 and those values are 0 or 1.)
If $X\in\text{up }\mathcal L_0\cap\text{up }\mathcal L_1$, then $X_0$ contains all but finitely many elements of $\mathbb N$ while $X_k$ for any $k>0$ contains both 0 and 1.  It follows immediately that $\prod_{i\in N}X_i$ contains an element of $f$, in fact infinitely many elements of $f$.
Now consider any fixed $c\in\{0,1\}^n$, and notice that there is at most one $k\in N$ such that the sequence $x=(k,c(1),c(2),\dots)$ (i.e., $c$ with $c(0)$ replaced by $k$) is in $f$.  Let $Q=U-\{k\}$ if there is such a $k$ and $Q=U$ otherwise.  Then one possibility for the $Y_i$ in (my interpretation of) the question is $Y_0=Q$ and $Y_i=\{c(i)\}$ for all $i>0$.  Then $\prod_{i\in N}Y_i$ does not contain any element of $f$.
