The W and Z gauge bosons receive radiative corrections in loop from the heavy SUSY scalars. There is an argument using gauge invariance which explains how the masses remains protected. I am not able to understand how gauge invariance is protecting the masses of W and Z.
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2 Answers
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Gauge invariance protects the W and Z boson masses even without supersymmetry. The W and Z bosons are the massive gauge bosons from spontaneously broken electroweak symmetry.
Consider the simpler case of an Abelian gauge symmetry. In this case, the symmetry acts on the gauge boson as a shift: $A_\mu(x) \to A_\mu(x) + \partial_\mu f(x)$ for some function $f(x)$. A mass term, $m_A^2A_\mu A^\mu$ is not invariant under this symmetry. Thus gauge invariance (as a shift symmetry here) prohibits a mass term for an unbroken Abelian gauge symmetry. This is true even at the quantum level: heavy particles that are charged under the gauge symmetry can contribute to the $A_\mu$ two-point function, but they only contribute to the wavefunction renormalization and cannot generate a mass term.
[ Remark: at this point, you can generalize to the transformation of a non-Abelian gauge boson like the $W$ or $Z$. The transformation is a little more complicated, but the conclusion is the same: the transformation rule of the gauge boson prohibits the mass term because the mass term would not be gauge invariant. ]
Now what happens when the gauge symmetry is broken spontaneously? Sticking to our simpler Abelian example, some field (a Higgs field) picks up a vacuum expectation value (vev) $v$. This is the order parameter of gauge symmetry breaking. The kinetic term of the Higgs field $|DH|^2$ includes interactions to the gauge boson through the covariant derivative, $D_\mu = \partial_\mu - ieA_\mu$. Plugging in $H = (h+v)/\sqrt{2}$ gives a mass term to the gauge boson: $\frac{1}{2}g^2v^2 A^2$. What has happened here?
The gauge boson now has a mass. The mass is proportional to the order parameter of gauge symmetry breaking, $v$.
Because we assume $v$ is the only order parameter of gauge symmetry breaking, any contribution to the mass of the gauge boson must be proportional to $v$.
This second point includes the first, but is more general. A loop of heavy particles contributing to the $A_\mu$ two-point function may now contribute to the mass of the gauge boson, but \emph{only} if it is proportional to $v$. That is to say: only if it contains an insertion of the Higgs vev. In fact, by (non-linearly realized) gauge invariance, the contribution to $m_A^2$ must be proportional to $\langle |H|^2\rangle = v^2$.
This shows how the gauge boson masses are protected, even when the gauge symmetry is broken. For any contribution to the gauge boson mass, we now have that $$\Delta m_{A}^2 \propto v^2$$ There's some prefactor; it certainly contains a $g^2$, but there may be additional couplings and loop factors. By dimensional analysis, then, the contribution to the gauge boson mass \emph{cannot} depend on a positive power of the cutoff, $\Lambda$, of the effective theory.
In other words, without the observation that (nonlinearly realized) gauge invariance requires two powers of the Higgs vev, one may have incorrectly thought that $$\Delta m_A^2 \propto \Lambda^2$$ from which we might think that $m_A$ should be on the order of the cutoff of the theory. However, because we know $\Delta m_A^2 \propto v^2$, we know from dimensional analysis that at most $\Delta m_A^2$ is logarithmically dependent on $\Lambda$: $$\Delta m_A^2 \propto v^2 \log(\Lambda/M)$$ where $M$ are the physical mass-scales of the contribution (e.g. masses of the particles in the loops).
One can generalize the above argument to a non-Abelian gauge symmetry that is spontaneously broken, as is the case for electroweak symmetry in the Standard Model. The above argument does not qualitatively change.
In a supersymmetric theory nothing changes. The gauge boson masses are still protected by gauge invariance. The (Majorana) gaugino masses inherit this protection due to supersymmetry.
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What I say below are very general facts and probably this is not the final answer you were looking for but maybe it helps.
A gauge theory (forget about SUSY for the moment) gives rise to a massless spectrum of gauge bosons and massless matter content. If you want to give mass to your gauge bosons you need spontaneous symmetry breaking terms in your lagrangian (this means the absolute minimum of your potential is not unique). Furthermore, if you want matter particles to be massive you need to add Yukawa terms to your lagrangian. Assuming there is no spontaneous symmetry breaking one says "the maslessness of the gauge bosons is protected by gauge invariance" because you an explicit mass term would violate gauge invariance.
If SUSY is present but is not broken then your spectrum will be richer but again as long as your gauge invariance is not broken there is no reason to expect massive gauge bosons neither gauginos.
Now, what happens when SUSY is there but you break gauge invariance?
What happens if you break both SUSY and gauge invariance?
I am sorry but I do not know the answer to any of these...it seems to me that if only gauge symmetry is broken your scalar fields (and superpartner) will pick up vacuum expectation values in such a way that all particles and super-particles have the same mass. So you have masses but they have to match.
In the second case I guess you will have massive spectrum but the masses of particles and superpartners will not match.
Sorry I was not more helpful :(
