Is the Ground State in QM Always Unique? Why?
I believe that it is true as long as there does not exist a non-trivial unitary operator $U$ that commutes with the Hamiltonian ($[H,U] = 0$) in the subspace of ground states. If such an operator exists then for a ground state $|\phi_0\rangle$ with energy $E_0$ we have $$HU|\phi_0\rangle = UH|\phi_0\rangle = E_0\left(U|\phi_0\rangle\right)$$ and so $U|\phi_0\rangle$ also has the lowest possible energy $E_0$ and it thus also a ground state. Note that the statement of non-triviality of $U$ is important. It needs to be non-trivial in the subspace of ground states, that is $U|\phi_0\rangle \neq e^{i\theta}|\phi_0\rangle$ for any phase $\theta$, otherwise there is no degeneracy. (Unitarity is needed so that $U|\phi_0\rangle$ is a state with norm 1)
More succinctly, if there exists a unitary operator $U$ such that $[H,U]=0$ and $U|\phi_0\rangle\neq e^{i\theta}|\phi_0\rangle$ for any phase $\theta$ then we have ground state degeneracy.
In the example you have given we see that the matrix elements in the basis given $\{|a\rangle,|b\rangle,|c\rangle\}$ is $$H = \begin{pmatrix}1&0&0\\0&1&0\\0&0&2\end{pmatrix}$$ from which we see there exists a unitary operator, with matrix elements $$U = \begin{pmatrix}0&1&0\\1&0&0\\0&0&1\end{pmatrix}$$ which commutes with $H$ and is non-trivial in the ground-state space.
Proof that non-existence of $U$ implies non-degenerate ground state:
Assume $\nexists U$ s.t. $\{[H,U]=0 ~~\mbox{and}~~ U|\phi_0\rangle \neq e^{i\theta}|\phi_0\rangle\}$
Now, for every state $|a\rangle$ and $|b\rangle$, $\exists U_{ab}$ which is unitary that takes us from $|a\rangle\rightarrow|b\rangle$. We are interested in the operator that take us from $|\phi_0\rangle$ to any $|a\rangle$ in our Hilbert space (which obviously includes all possible ground states), which we denote by $U_{a0}$. This means that any state $|a\rangle$ can be written as $|a\rangle = U_{a0}|\phi_0\rangle$. By our starting assumption $U_{a0}$ either satisfies $$(1)~~~~~~ [H,U_{a0}]\neq 0,~~~~~~~~\mbox{or}~~~~~~~~(2)~~~~~U_{a0}|\phi_0\rangle = e^{i\theta}|\phi_0\rangle$$ If (1), then we have $$H|a\rangle = H U_{a0}|\phi_0\rangle \neq U_{a0}H|\phi_0\rangle = E_0|a\rangle~~~\implies~~~H|a\rangle \neq E_0|a\rangle$$ and so $|a\rangle \neq |\phi_0\rangle$ is not a ground state.
If (2), then $|a\rangle = e^{i\theta}|\phi_0\rangle$ and so $|a\rangle$ and $|\phi_0\rangle$ represent the same state.
Thus the non-existence of $U$ implies the non-existence of a second ground state and thus non-degeneracy.
To be clear:
Is the ground state of a quantum system always non-degenerate?
the answer is an unequivocal no. Real quantum systems can and do have degenerate ground states.
Some examples:
For a three-level system with hamiltonian $$H=\begin{pmatrix}1& 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2\end{pmatrix},$$ as given by the OP, the ground state is degenerate. This should be all that's necessary to show that the claim is false in general.
Pretty much all atoms in a field-free vacuum have degenerate ground states, with the simplest examples being boron and carbon, which have $p$-shell electrons that fit multiple orthogonal magnetic-quantum-number states at exactly the same energy. The same is true for pretty much all of the periodic table, with the exception of atoms with full subshells. Thus, the alkaline earth metals, the noble gases, and the rightmost columns of the transition metals and the rare earths, have nondegenerate ground states, and everything else is degenerate.
(On the other hand, it is important to note that these kinds of degenerate ground states can be relatively fragile, so e.g. if the atom wanders into a stray bit of magnetic field, that will lift the degeneracy, often by a nontrivial amount. However, that doesn't mean that the free-atom ground state isn't degenerate.)
This is exactly the same situation as that pointed out in a comment, regarding atomic nuclei, whose ground state will generically have nonzero angular momentum and will therefore be spatially degenerate.
A slew of ferromagnetic and anti-ferromagnetic materials in lattices that exhibit geometrical frustration, which is best exhibited graphically:
That is, if three spins are linked with pairwise anti-ferromagnetic couplings, they try to point in opposite directions to each other, but there's no global solution that will avoid high-energy parallel alignments. This then leads naturally to a degenerate ground-state manifold.
Now, there is a large class of hamiltonians for which the ground state can be shown to be nondegenerate ─ they're explored in some depth, and with good references, in this MathOverflow thread ─ which includes many hamiltonians of the form $-\nabla^2 +V$, regardless of the dimension, for distinguishable quantum particles. However, but this class does not include all possible systems, particularly once you include fermionic particle statistics with strict antisymmetry requirements on the wavefunction.