**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
 $$ \exists c \in \{ 0, 1 \}^N \,
\forall Y\in \prod\_{i\in N}\mathcal{L}\_{c (i)} (i) :
\left( \prod\_{i \in N} Y\_i \right) \cap f \neq \emptyset . $$



Interweaving two indexed families of filters
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 $f$.


