I think this is a very good and deep question that maybe related to Prof.Wen's question, and I would try to answer you in my understanding.
Let's take the nearest-neighbor spin-1/2 antiferromagnetic Heisenberg Model on the square lattice as an example, where the symmetry breaking Neel state only emerges in the thermodynamic limit.
As you have mentioned, note that only when the system has finite size, (i.e., the square lattice constitutes of two sublattices A and B with equal sizes $N_A=N_B$, and hence the total number of spins $N=N_A+N_B$ is even), the ground state is unique and is exact a singlet state with $SU(2)$ spin-rotation symmetry. However, as the system size becomes large, there are many low-lying excited states with very small gap $\Delta$ above the singlet ground state, and these low-lying states break the $SU(2)$ spin-rotation symmetry(i.e., they may be triplet states). More subtlely, as $N$ approaches $\infty $, those nearly degenerate ground states would 'collapse' into the ground state in the thermodynamic limit, indicating that the Neel state is in fact a superposition of many nearly degenerate ground states in the thermodynamic limit. Thus, in the strict thermodynamic limit, there exists an $SU(2)$ symmetry breaking state of the 'highly' degenerate ground states.
The above is just an very rough picture, and I also feel a little hard to understand the ground state degeneracy for a gapless system in the thermodynamic limit. Moreover, I also get a question: theoretically, as there are 'highly' degenerate ground states containing both $SU(2)$ symmetric and symmtery-breaking Neel states in the thermodynamic limit, why we used to say the ground state of antiferromagnetic Heisenberg Model on the square lattice is a Neel state rather than an $SU(2)$ symmetric singlet state?