What is the link between the BCS ground state and superconductivity?
An electron-excitation-gap is necessary helpful for superconductivity. (Normally electrons scatter, but if there's a gap then they don't scatter, because there's no state to scatter into, as long as the temperature is low enough that they cannot jump the gap.**) But it is not sufficient. The filled states also have to be able to carry a current! The electrons in a filled semiconductor valence band have an electron-excitation-gap, like you say, but they do not carry a current. (If you think about it, in intrinsic GaAs near absolute zero, there are no electron scattering events!) On the other hand, the filled states in a superconductor CAN carry a current because...well I wasn't sure but I read the old BCS paper and they have a pretty basic explanation:
Our theory also accounts in a qualitative way for those aspects of superconductivity associated with infinite conductivity...the paired states $(k_{1\uparrow}, k_{2\downarrow})$ have a net momentum $k_1+k_2=q$, where $q$ is the same for all virtual pairs. For each value of $q$, there is a metastable state with a minimum in free energy and a unique current density. Scattering of individual electrons will not change the value of $q$ common to virtual pair states, and so can only produce fluctuations about the current determined by $q$." And these scattering events raise the free energy, unless all of the electrons scatter simultaneously in exactly the right way to make a new metastable state centered at a different q, which is extremely unlikely.
Well that makes sense to me.... In your question you wrote the BCS ground state with $q=0$, but that's just one of the family of BCS meta-stable (ground-ish) states with different $q$, with different $q$s corresponding to different current flows.
In other words, BCS theory explains how the electrons pair up and then there is an energy gap for single-particle-excitations. Slightly changing the $q$ requires little or no energy (or in some cases even lowers the energy), but will not happen spontaneously because it requires trillions of electrons to change their state simultaneously in a coordinated way. (An electric field can cause this kind of coordinated change, but it can't just happen spontaneously. Generally, only single-particle-excitations happen spontaneously, and these are gapped.) So it is metastable. And the fact that you can have a metastable state which carries current is just another way of saying that current can keep flowing and flowing even with no electric field pushing it.
**Update: OK, yes there is such a thing as "gapless superconductivity". My mistake was conflating "superconductor" with "superconductor with no dissipation whatsoever". The latter doesn't exist—even with a full proper superconducting energy gap, remember the superconducting transition is above absolute zero, so there is certain to be some small but nonzero electron scattering rate that is tolerable without destroying the superconducting order. So by that logic, it's not surprising that a partial or nonexistent gap is compatible with superconductivity at a very low transition temperature.