Are Turing-recognizable languages closed under intersection?
Indeed, the recursively enumerable languages are closed under union and intersection, but not complement. This is not a contradiction: the formula $A \cap B = \overline{\overline{A} \cup \overline{B}}$ tells that a class closed under union and complement is closed under intersection, but does not tell anything otherwise.

A similar example: the set of finite subsets of $\mathbb N$ is clearly closed under union and intersection, but not complement.