Any simple reason why Helium in the ground state is diamagnetic? I know the electrons are in the spin singlet state, and the spatial part of the wave function is an S-state. 
But that is not sufficient for it to be diamagnetic. 
 A: Since the two electrons are in the 1s state, they must have opposite spins according to the Pauli exclusion principle.  
For helium-4 the nucleus has no spin, so it does not contribute.  
For helium-3, the nucleus is spin 1/2 and make a small paramagnetic contribution, so helium-3 is less diamagnetic than helium-4.
Dimerization of helium also has a small effect. 
See Diamagnetism of helium and references cited therein for further information.
A: The interaction with a magnetic field leads partial p-character of the electrons. This makes the electrons to actually have non-zero angular momentum so that they do interact with an external magnetic field, or in other words you can actually induce a ring current with these electrons in an external field:
https://iopscience.iop.org/article/10.1088/0953-4075/33/3/322
A: I think what he wanted to ask is why diamagnetism is observed for Helium if the $1\mathrm s$ orbital has an orbital angular momentum of $0$.
Orbital angular momentum depends on the $m\ell$ quantum number, which is $0$ for the $\mathrm s$ states.
And Langevin diamagnetism is explained by orbital angular momentum precessions, but both $1\mathrm s$ electrons in Helium appear to not have an orbital angular momentum.
I think the explanation here is a bit more complicated, and has to do with the physical significance of the orbital angular momentum quantity, whether it really means it's $0$, or it simply is averaged as $0$, since for a spherical wavefunction the orbital angular momentum would be quenched regardless of its actual value.
Edit: ok i might have forgotten a minor detail that is even more important, the mL quantum number itself doesn't give the value of actual magnetic moment, only a projection on the Z axis, the total magnetic moment depends on the l quantum number.
The s orbitals have the l quantum number equal to 0, the l quantum number allows you to calculate the value of the magnetic moment.
μL=sqrt[l(l+1)]μB   
so for l=0 this would give a 0 value of magnetic moment, now i myself am not sure wether this simply means it's a quenched magnetic moment or it actually IS 0 at all times no matter what. Because if it is 0, then it would be impossible for an s orbital to have any contribution to diamagnetism, since that requires the magnetic moment to be able to precess around an external field such that it produces an opposite response. 
Someone with a better understanding of magnetic moments in QM should clarify this.
