It is true both that the system has an infinite number of ground states, and that the ground state is $(2LS+1)$-fold degenerate, and there is no contradiction.
That is, in the $(2LS+1)$-dimensional space of ground states there are an infinite number of physically distinct states, each corresponding to a choice of the direction in which all of the spins are pointing, but these states are mostly not orthogonal to each other.
If you think about a single spin-$\frac12$ particle, this is already the case - there are an infinite number of states, but they form a finite dimensional space.
The Heisenberg model is actually an example of an exception to the standard Goldstone theorem for relativistic QFT. In the standard case, we expect that each broken symmetry yields a gapless mode with linear dispersion at small momenta. This is not true generally for non-relativistic systems such as the Heisenberg model. Consider the Hamiltonian
\begin{equation}
H=-J\sum_{n=1}^N\vec{s}_n\cdot\vec{s}_{n+1},
\end{equation}
where we assume periodic boundary conditions $\vec{s}_{N+1}=\vec{s}_1$. Here we are describing a spin chain rather than a lattice for simplicity. Since the Hamiltonian couples nearest neighbors through a dot-product, the Hamiltonian itself exhibits rotational symmetry about the $x$, $y$, and $z$ axes. The ground state of this system has all the spins pointing in the same direction, which we will choose to be the $+z$ direction; that is,
\begin{equation}
|0\rangle=|\uparrow,...,\uparrow\rangle.
\end{equation}
Thus the vacuum, or ground state, of the theory has chosen a preferred direction in space and has broken the original rotational symmetry. Rather, $|0\rangle$ is only invariant under rotations about the $z$-axis. You can verify this fact for yourself by acting on the ground state with the standard SU(2) rotation operators (one for each lattice site). We say that the original SO(3) symmetry has been broken to SO(2). We have two broken symmetries in this case, corresponding to rotations about the $x$ and $y$ axes.
Now consider the following state
\begin{equation}
|k\rangle=\sum_{n=1}^N e^{ikn}|...,\uparrow,\downarrow_n,\uparrow,...\rangle,
\end{equation}
which is a linear combination of states with the $n$-th spin flipped down. One may show that this state is an energy eigenstate with excitation energy (above the ground state)
\begin{equation}
\Delta_E=\hbar^2J(1-\cos(k)),
\end{equation}
which is gapless and quadratic in $k$ for small $k$. This excitation is known as a magnon. The curious thing here is that we broke two symmetries but found only a single excitation with quadratic rather than linear dispersion. The reason is that the two symmetries that we broke are not independent since the generators of those symmetries satisfy the familiar commutation relations
\begin{equation}
[\sigma_i,\sigma_j]=2i\epsilon_{ijk}\sigma_k,
\end{equation}
so that in particular
\begin{equation}
[\sigma_x,\sigma_y]=2i\sigma_z.
\end{equation}
By performing rotations about the $x$ and $y$ axes, we can generate a rotation about the $z$-axis.
The number of broken generators is exactly equal to the number of Goldstone bosons if
\begin{equation}
\langle 0|[Q_i,Q_j]0\rangle =0,
\end{equation}
for all broken symmetry generators $Q$. Now you may wonder what is so special about Lorentz invariance. Recall that Noether's theorem tells us that conserved charges (i.e. symmetry generators) can be written as the integral of a conserved current
\begin{equation}
Q=\int j^0(x) d^3x,
\end{equation}
whereas Lorentz invariance demands that expectation values of objects with free Lorentz indices must vanish
\begin{equation}
\langle j^0(x)\rangle=0,
\end{equation}
thereby ensuring that the number of Goldstone bosons is always equal to the number of broken symmetry generators in a relativistic theory.
You may find the following reference useful (the source of most of this info).
Spontaneous Symmetry Breaking
Best Answer
You are correct that Goldstone's theorem does not apply to the 1-D Heisenberg chain, or indeed to any local 1-D system, because there is no continuous symmetry breaking in 1-D for local Hamiltonians. But Goldstone modes are not the only possible kinds of gapless excitations - you don't need continuous SSB for a system to be gapless. There are plenty of examples (other than the Heisenberg chain) of gapless systems that don't break any continuous symmetries, like critical systems, algebraic spin liquids, or systems with Fermi surfaces.
Put another way, Goldstone's theorem says that continuous SSB implies gapless excitations, but the converse of Goldstone's theorem is not true: gapless excitations do not imply continuous SSB.
The low-energy excitations of the antiferromagnetic Heisenberg chain are "spinons," which can be roughly thought of as the free spins that arise from breaking spin singlets in the ground state. Their gaplessness can be proven using the Bethe ansatz, or by using the fact that the emergent low-energy QFT description is a Wess-Zumino-Witten model, which is a CFT and therefore gapless, as an energy gap would set a scale and break conformal invariance.