Is sound a Nambu-Goldstone mode? The usual sound exists in solids, liquids, and gases, as a long-wavelength excitation with linear dispersion. Can its presence be attributed to the spontaneous breaking of some symmetry? In other words, is it a Goldstone mode of some symmetry?
 A: It depends on what sound you are talking about. Yes, in crystalline solids, sound is nothing but propagating phonons, in which case it is the Goldstone modes corresponding to broken translational symmetry.
In fluids, sound is a pressure-density wave and there it is not a Goldstone mode. It instead arises because of conservation laws governing the conservation of momentum and of mass (non-relativistically and in the absence of reactions) along with the presence of inertia.
A: Yes sound is a goldstone mode. Consider, for example, an ideal gas with particles at positions $\mathbf{x}_i$. There is a symmetry where we can displace each particle by some displacement $\mathbf{u}$. Of course this symmetry breaks spontaneously. By definition, we only observe $\mathbf{u}=\mathbf{0}$.
The goldstone modes corresponding to this symmetry are modes where $\mathbf{u}$ is nonzero and varies spatially with some wavevector $\mathbf{k}$. That is, each particle gets displaced according to $\mathbf{x}_i \to \mathbf{x}_i + \mathbf{u} \cos(\mathbf{k} \cdot \mathbf{x}_i)$. This displacment will cause a sinusoidal variation in density, and therefore  sinusoidal varation in pressure, which is what sound is. 
Notice that the energy of the mode goes to zero as $\mathbf{k}$ goes to zero, since the $\mathbf{k}=\mathbf{0}$ limit is just a uniform shift, which requires zero energy. That is the idea behind goldstone modes. This same logic applies in liquids and solids as well.
