The Clique vs. Independent Set Problem
The two players construct a sequence $V_0 \supset V_1 \supset \cdots \supset V_m$ of sets of vertices such that:


$V_0$ consists of all vertices in the graph.
$|V_{i+1}| \leq (|V_i|+1)/2$.
$V_i \supseteq C \cap I$.


The players stop once $|V_m| \leq 1$. At this point they can answer the question using $O(1)$ communication.

At round $i$, the players know $V_{i-1}$, and wish to construct $V_i$. They act as follows:


If $C \cap V_{i-1}$ contains a vertex $v$ with fewer than $|V_{i-1}|/2$ neighbors in $V_{i-1}$, then Alice sends Bob one such vertex $v$, and both players set $V_i$ to be this set of neighbors, together with $v$ (this is valid since $C \cap I \subseteq C \subseteq V_i$). Otherwise, she sends $\bot$.
If Alice sent $\bot$ and $I \cap V_{i-1}$ contains a vertex $v$ with fewer than $|V_{i-1}|/2$ non-neighbors in $V_{i-1}$, then Bob sends Alice one such vertex $v$, and both players set $V_i$ to be this set of non-neighbors, together with $v$ (this is valid since $C \cap I \subseteq I \subseteq V_i$). Otherwise, he sends Alice $\bot$.
If both players sent $\bot$, then $C \cap I = \emptyset$. Indeed, if $v \in C \cap I$, then $v$ has at least $|V_{i-1}|/2$ neighbors and at least $|V_{i-1}|/2$ non-neighbors inside $|V_{i-1}|$, whereas the number of potential neighbors and non-neighbors is just $|V_{i-1}|-1$. Therefore the players can abort and conclude that $C \cap I = \emptyset$.


Each round takes $O(\log n)$ bits of communication, and there are $O(\log n)$ rounds, for a total of $O(\log^2 n)$ bits of communication.