How does gauge invariance prevent the photon from acquiring a mass? I have read that the photon is protected from acquiring a mass due to gauge invariance. However, I was under the impression that gauge invariance is not a true physical symmetry, but rather a redundancy in description of a given physical theory? 
Is it simply that gauge invariance is required such that one can remove the two unphysical degrees of freedom in the photon field. If the theory weren't gauge invariant then one would not be able to do this. Accordingly, requiring gauge invariance prevents one from writing down a mass term for the photon field sinc such a term is not gauge invariant?!
If this is true, however, I have read that without the Higgs mechanism all standard model particles are massless and that this is due to the gauge symmetry, so what is the general reason for why this is the case if gauge symmetry is simply a redundancy in our description?
 A: 
However, I was under the impression that gauge invariance is not a
  true physical symmetry, but rather a redundancy in description of a
  given physical theory?!

This is true. Instead of the symmetry transformation
$$
U|\Psi\rangle \to |\Psi'\rangle
$$ 
the unitary gauge transformation operator $U$ with the generator $G$ acts on physical state $|\Psi\rangle$ as
$$
\tag 1 U(G)|\Psi\rangle = |\Psi\rangle \leftrightarrow G(\mathbf x)|\Psi\rangle = 0
$$
Analogically, in theory with interactions there is another "do-nothing transformation", called the Ward identity. It states that any amplitude $M$ with at least one external photon line, i.e., $M = M_{\mu}\epsilon^{\mu}(p)$, is transversal:
$$
\tag 2 p^{\mu}M_{\mu} = 0
$$
It is the analog of gauge current conservation in the classical theory. 
Together, $(1)$ and $(2)$ can be called as "the gauge invariance of the quantum field theory", but in fact they are just the description of the gauge redundancy.
It turns out that typically $(2)$ forbids the mass of the photon. Precisely, it can be shown that in most cases it prevents the position of the photon propagator pole (which defines its mass) to be shifted from zero.
However, in some particular cases we can write down the gauge field mass without violating the gauge invariance (this statement in fact is not exact, see below, however typically people say it as "mantra").
Really, classically one can write down the gauge-invariant field
$$
\tag 3 V_{\mu} = A_{\mu} - \partial_{\mu}\frac{1}{\square}\partial_{\nu}A^{\nu}
$$ 
Corresponding mass term $m^{2}V_{\mu}V^{\mu}$ doesn't violate the gauge invariance, although it looks cumbersome. The second summand in $(3)$ looks bad, but the situation becomes good when there is some scalar degree of freedom $\varphi$ which parametrizes it:
$$
\frac{1}{\square}\partial_{\nu}A^{\nu} \to \varphi
$$
In the quantum field theory corresponding situation appears when the photon vacuum polarization amplitude has the pole at zero momentum.
There are few models in which such situation appears. The first one is the model with the Higgs mechanism (often called "spontaneous breaking" of the gauge symmetry), where the particles spectrum doesn't form the representation of the gauge group below some scale $\Lambda$. Instead of $A_{\mu}$ we're dealing with $V_{\mu}$; people say that $A_{\mu}$ eats the "Goldstone boson" $\varphi$. But note that in fact $V_{\mu}$ is not the "photon"; it is the combination of the photon and longitudinal polarization. 
The prototype of this idea is the classical theory of EM field interacting with the plasma (this was pioneered by Anderson).
The second example is strong interaction regime bozonization. It happens in Schwinger's massless 2D QED. In this case, two massless fermions form the massless bound state shifting the photon propagator pole. One can't write down the local term generating the photon mass without introducing the new field representing this bound state.
