I have been looking at the following problem
Suppose $k,l \geq 1$, $k+l \leq n$ and consider $$\mathcal{A} \subseteq \binom {[n]} l, \mathcal{B} \subseteq \binom {[n]} k$$ Which are cross intersecting, that is if $A \in \mathcal{A}, B \in \mathcal{B}$, then $A \cap B \neq \emptyset$.
Then we must have that either $\left| \mathcal{A}\right| \leq \binom {n-1}{l-1}$ or $\left| \mathcal{B}\right| \leq \binom {n-1}{k-1}$.
This look like a simple application of Erdős–Ko–Rado, but so far I am having problems seeing it. Any help is appreciated (hints preferred).
I don't see how to apply Erdős-Ko-Rado, but you can modify Katona's double-counting proof of EKR to show that, given your hypotheses, $$|{\cal A}||{\cal B}|\le {n-1\choose l-1}{n-1\choose k-1}.$$
The hint is to count quadruples $(C_A, C_B, S, T)$ where $C_A$ and $C_B$ are cyclic orderings of $[n]$, $S\in\cal A$ is an $l$-interval under $C_A$, and $T\in\cal B$ is a $k$-interval under $C_B$.