Why symmetry breaking? Let me elaborate the question by using 2D Ising model without external magnetic field. When we lower the temperature and pass $T_c$ a little bit, the theory of spontaneous symmetry breaking tells us that the system will settle in one of the two degenerate ground states with non-zero magnetization.
My question is why cannot the system be in a superposition of the two states and thus the average magnetization still remains zero. People may say that at low $T$ the system does not have enough energy to tunnel through the barrier in the middle of the double-well free energy landscape. Well, this might be true for relatively high barrier, but when $T\rightarrow T_c^-$ the barrier can have a height that's close to 0, in which case the superposition of states seems to be possible and the average magnetization should be zero if that is so, which in turn contradicts the results from symmetry breaking.
Is it because the system is supposed to be coupled with a heat bath such that quantum decoherence destroys the superposition? But What about spontaneous symmetry breaking in quantum phase transitions?  
 A: This is an extremely deep question that still isn't fully understood. Such a "macroscopic superposition" $|\uparrow \uparrow \uparrow \uparrow \dots \rangle + |\downarrow \downarrow \downarrow \downarrow \dots \rangle$ is a perfectly valid state in the Hilbert space, and yet we never see it experimentally (at least not without a lot of careful work to isolate the system). Such a state is called a "cat state" (after Schrodinger's cat) and you are essentially asking about the paradox of why we never see Schrodinger's cats walking around.
More precisely, a cat state is a state $|\psi\rangle$ that fails to obey the "cluster decomposition property," which says that for all local operators $\hat{A}(x)$ and $\hat{B}(y)$,
$$\lim_{x - y \to \infty} \left[ \langle \psi | \hat{A}(x) \hat{B}(y) | \psi \rangle - \langle \psi | \hat{A}(x) | \psi \rangle \langle \psi | \hat{B}(y) | \psi \rangle \right] = 0,$$
i.e. expectation values of far-separated observables factorize. You can easily check that your proposed superposition does not satisfy the cluster property, while the physically realistic symmetry-broken states do.
The most widely accepted resolution is that as you suggest, cat states are unstable to decoherence. Before you measure the system, it can indeed be in such a macroscopic superposition. But when you do measure it, the measuring device itself acts as a thermal bath that produces the random jitters that destroy the coherent superposition! A measuring device is always macroscopic and can act as an entanglement bath. (Indeed, under the current way of thinking, the property of being macroscopic is what makes it a measuring device! Of course, a "measuring device" doesn't need to have a conscious human watching it or anything like that - random air molecules bouncing off the system and so on can "measure" the system and collapse/decohere any macroscopic superpositions.)
A: Ising model is at his core a classical model. Meaning that we don't consider the possibility of a "superposition of states" in the quantum sense. 
If you had an experiment recreating the Ising model (a sort of magnet, I guess), that kind of quantum effect would appear. But a model is not an experiment and we voluntary don't take into account quantum effects.
But, "Quantum" Ising model do exists, and have exactly the sort of behaviour you seems to expect, if there is multiple ground states possibles, the system will be in a superposition of states. (see by example https://en.wikipedia.org/wiki/Heisenberg_model_(quantum))
As for symmetry breaking, it's still here even in the quantum world, as soon as you measure the state, you will fall in one of the ground state.
