Why is the electromagnetic field massless?
I presume the mass term in the Lagrangian, $$ \dfrac{1}{2}M^2 A_{\mu}A^{\mu}, $$
would "violate" gauge redundancy.
But this is what its name indicates: redundancy!
Who cares if it's lost? What are the consequences of losing gauge redundancy in order of importance?
p.s.: On a more serious note: how does one improve his questions?
You can have massive vector fields. These fields will not have gauge-invariance. This is not a problem, we didn't ask for gauge invariance when constructing the theory of electromagnetism. The problem is that massive fields mediate interactions that fall off as $r^\# e^{-mr}$, where $m$ is the mass of the field, and $r$ is the distance. Because of the exponential fall-off, the range of the force will be $1/m$.
For massless photons the force falls off as $1/r^2$, making it long range.
Since we know static electromagnetism has rather long range (EM waves are not relevant here -- these correspond to real photons, while force is being mediated by virtual photons), we are pretty sure that photon is massless. Furthermore, if photon was massive, the speed of light would depend on the reference system. You can find some discussion and links regarding bounds on photon mass on this wiki page. Particle data group gives a list of upper bounds on photon mass, with their accepted upper bound being $10^{-18}$ eV. (The bound quoted on wikipedia page is debated.) This corresponds to range of force of at least $10^{11}$ meters. This is the same order of magnitude as distance to the Sun.
Gauge symmetry is a redundancy in the sense that it stems from the fact that we describe an object with more degrees of freedom than those that are actually needed. That is $A_\mu$ and $A_\mu + \partial_\mu \phi$ describe the same physical object, and thus any total derivative contribution to $A_\mu$ is redundant.
Now the conclusion that must be drawn from this fact is precisely the opposite. Since gauge symmetry is a redundancy than it is sacred! If it's broken, i.e. $A_\mu$ and $A_\mu + \partial_\mu \phi$ give different physical consequences, it means that the same physical object behaves differently if we describe it differently on paper. This is of course absurd.
Moreover, $A_\mu$ is never measured, but $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$ is. So after measuring a certain value of $F$ it better be that, no matter the choice we make for writing down an $A$, the result is the same.
