Solution: Given that $A$ is countable and $B$ is uncountable. We claim that $B\setminus A$ is uncountable. For that, let us suppose that $B\setminus A$ is countable, then we have \[ B = \left( B\setminus A \right) \cup A. \] Since $A$ is countable and $B\setminus A$ is supposed to be countable, and union of countable sets is countable and hence $B$ is countable, a contradiction. Therefore, $B\setminus A$ is uncountable.