Whether it is necessary to search still for variants of an explanation of spontaneously breaking gauge symmetry, giving masses for a W, Z-bosons?
Goldstone bosons are bosons that appear necessarily in models exhibiting spontaneous breakdown of continuous symmetries, thus it is clear that the Higgs bogon - is a Goldstone boson. If LHC searches of a Higgs boson won't be a success, whether it will mean a theorem inconsistency for electroweak interactions?
The broken gauge symmetry model for the W and Z boson masses is correct with scientific standards of certainty, since it is exactly verified by relating the ratio of their masses to the ratio of the coupling constants, and there is no chance for such a precise agreement by accident. The pattern of breaking tells you that there is a charged condensate in the vacuum of the form of the usual Higgs, a scalar SU(2) doublet ("spin 1/2") representation with U(1) charge 1/2 (the same as the lepton doublet). But there is no real reason to think that this must be a fundamental scalar field with these charges.
It is good to have as wide a range of models for the Higgs sector as possible. It is not possible for this field to just what we have seen so far, just the transverse modes of the W and Z, because these modes can't be unitary all by themselves. Their dynamics is that of a particular nonlinear sigma model, the limit as the self-coupling $\lambda$ goes to infinity of the standard model Higgs, and this limit is inconsistent (see addendum). So you need something to unitarize in the absence of something like the Higgs, a good dynamical field theory which reprouces the Higgs condensate.
So it is just inconceivable that the LHC will find nothing at all. Once you get data on the missing component(s) of the Higgs, it will reveal if the higgs condensate is a composite fermion condensate, as in technicolor, or a scalar condensate as in the standard model and SUSY variants, or something else entirely (like a composite scalar made out of technicolor scalars or something even more exotic like an infraparticle of a higher energy Banks-Zaks theory or whatever).
The Goldstone theorem just doesn't work when there are gauge fields coupled to the symmetry, this is the whole point of the Higgs mechanism. The long-range Coulomb interaction coupled to a charged condensate produces no Goldstone bosons. The argument is summarized on Wikipedia in the page on the Higgs mechanism, in the section on superconductivity. The mechanism is often called the "eating" of the Goldstone bosons by the Gauge bosons. It is no more mysterious than the statement that plasma waves have a finite frequency at infinite wavelength, because of the instantaneous coulomb repulsion (which is still relativistically instantaneous in Dirac gauge).
Like many field theory arguments, good guidance is provided by the Ising model. The Ising model is the $\lambda$ goes to infinity version of the scalar with field potential $\lambda(\phi^2 - 1)^2$. In the large $\lambda$ limit, when you discretize the action on a lattice, you force the field to be $\pm 1$. The coupling between neighboring field values reproduces an Ising model action of some kind (perhaps with next-to-nearest neighbor coupling, or whatever, it's the same universality class).
When you look at this statistical theory at long distances, you reduce to a $\phi^4$ theory with a $\lambda$ which goes down with larger distances. If you back-trace the evolution of $\lambda$, it gets stronger at shorter distances, and it blows up at the order of magnitude of the lattice scale (this is completely obvious, because the lattice scale is exactly where $\lambda$ is infinite, so that the Ising description is correct. Averaging over blocks of spins allows the field to fluctuate away from plus or minus 1, so it reduces the $\lambda$. This is also formally well known from the $\beta$ function calculation in $\phi^4$ theory).
So the bare Ising model lattice scale is where the coupling blows up, and we know by construction that there are no shorter distances consistently defined in this model. The lesson learned is that any scalar theory must have some new physics at the scale of its Landau pole (the scale where the coupling blows up), otherwise, you will see the lattice, or whatever structure is hiding behind the long-wavelength quantum/statistical field theory.
The nonlinear sigma model defined by the classical large $\lambda$ limit of the standard model can be defined at some microscopic scale, but then at long distances, it will flow to the standard model scalar Higgs with a weak coupling. If the cutoff is much larger than the TeV Higgs scale, the coupling is bounded above by this triviality argument and the constraint that the location of the Landau pole is higher than the cutoff scale. This gives Weinberg's Higgs mass bound. The reason it's a mass bound and not a coupling bound is because the Higgs particle mass is determined by the curvature of the Higgs potential in the hard direction of the Mexican hat (the soft directions are the Goldstone bosons that are eaten by the W an Z), and the curvature in this hard direction is proportional to $\lambda$.
This is called the unitarity bound, or the triviality bound, depending on who is speaking, and it certainly excludes an infinite coupling at TeV scales, as required for a nonlinear sigma model which would keep only the longitudinal models of the W and Z, and discard the Higgs boson by moving its mass to infinity.
Note that this argument doesn't work for Abelian Higgs mechanism, when the Higgsing is of a (noncompact) U(1) gauge theory, because you can take the Higgs charge to zero and the Higgs self-coupling to infinity and the condensate value to infinity while keeping the mass of the U(1) photon finite, and the nonlinear sigma model limit (just a circle) is the Stueckelberg Affine Higgs mechanism. This limit evades the triviality argument because the circle becomes big at the same time as the coupling becomes big. This doesn't work in the nonabelian case, because the charges are quantized with a lower bound.
This addendum is provided because I didn't feel comfortable just saying "Weinberg says so." But if it is too telegraphic, then, well, Weinberg says so.
