Computing the union closure
The complexity of enumerating maximal independent sets in graphs is the same as in bipartite graphs, so bipartiteness does not bring anything new. 

You have an algorithm (with exponential space) in $O(|C|\cdot n^2)$, but no polynomial space algorithm that acheives this time complexity is known. The following paper http://www.sciencedirect.com/science/article/pii/S0166218X08004563 is a good survey.