In the literature I read that the Hall conductivity is quantized because the Hall conductivity is actually the winding number associated with the mapping from the brillouin zone (a torus) to the space of Hamiltoninans. The calculation assumes that system has an energy gap above the ground state. But doesn't that mean that the system is an insulator ? If the last statement is true how does the system conduct ?
Yes. According to the classification scheme, the quantum Hall insulator is first an insulator, then a band insulator, then a topological band insulator (or abbreviated as a topological insulator), and finally a 2D topological insulator in symmetry class A.
But quantum Hall insulator can conduct Hall current! This statement is not a contradiction.
You need to understand the precise definition of insulators. A state of matter is classified as an insulator if the state does not conduct when it is placed on a closed manifold (manifold with no boundary). So the definition does not rule out the possibility that an insulator can conduct when it is placed on an open manifold (manifold that has boundary). In fact, the key feature of a topological insulator is insulating in the bulk and conducting on the boundary.
Quantum Hall insulator is indeed insulating and has no Hall conductivity if there is no boundary. One you cut open the boundary, gapless edge modes will emerge on the boundary, which then carry the Hall current and contribute to the Hall conductivity.