TL;DR:
If your Hamiltonian has a continuous symmetry, you can take the ground state and rotate it by an arbitrarily small amount, and it is still a ground state. If you patch together two blocks which differ by a small rotation, you only pay a price at the boundary between the regions, and this penalty can become arbitrarily small as you make the rotation smaller and smaller.
You can then gradually change this small rotation throughout your system, and the price you pay is only the change in rotation, which can be made arbitrarily small. (Symmetry breaking ensures that this state is different from the ground state; otherwise, you might just construct a state which is equal to the ground state.)
However (fortunately? unfortunately?), this is only half the story, since you have to add up these small energy penalties over a large number of positions (since you need to complete a full $2\pi$ rotation globally), and there is no guarantee that the sum of many small things will be small.
To show that this sum is indeed small, you also have to use that the ground state has an extremality property (namely that it is the ground state), which implies that the dependence of the energy on the rotation angle must be quadratic in the angle (a linear term would increase the energy in one direction, but it would necessarily decrease the energy in the other direction), while the number of terms is linear in the inverse angle, so the sum goes to zero for a large system (and thus small angles) at least as one over the system size.
What is different is the symmetry is discrete? For a discrete symmetry, a small rotation of the ground state gives a state which is no longer a ground state, so when you patch together two regions which differ by a small rotation, you pay a price proportional to the volume of one of the regions. On the other hand, if you patch together two regions related by a discrete symmetry transformation, the price you pay at the boundary is constant (as the angle is large). Thus, the excitations you can construct this way have a constant energy.
Consider a 1D Hamiltonian with a continuous symmetry $U_g$,
$$
U_g^{\otimes N}H(U_g^\dagger)^{\otimes N} = H\ .
$$
Then, if $\vert\psi\rangle$ is a ground state of $H$, so is $U_g^{\otimes N}\vert\psi\rangle$.
Now split the chain into two parts A and B, and apply $U_g$ to the sites in A, and $U_h$ to the sites in B. Then, the energy for the Hamiltonian terms inside A and B will still be the same. However, the terms at the boundary will have some higher energy. Assuming a two-body Hamiltonian $H=\sum h$, and denoting the two-site reduced density matrix by $\rho$, this energy will be
$$
e'=\mathrm{tr}[h(U_g\otimes U_h)\rho(U_g\otimes U_h)^\dagger] =
\mathrm{tr}[h(I\otimes U_{g^{-1}h})\rho(I\otimes U_{g^{-1}h})^\dagger]\ ,
$$
where I have used the symmetry of the Hamiltonian.
Now if the group is continuous, you can choose $g^{-1}h$ such that
$$
\|U_{g^{-1}h}-I\|\le \epsilon\ .
$$
This is what is special about the continuous symmetry, and what fails for a discrete symmetry!
Then, you will find that the change in energy at each boundary is
$$
\Delta e = e' - \mathrm{tr}[h\rho] \le 2 \|h\|\epsilon\ .\tag{1}
$$
You might now think that we are done, since we can now proceed to combine this by splitting the system into $L$ blocks which you rotate by $2\pi/L$ each. Then, $\epsilon \propto 1/L\to 0$. However, the problem is that you are adding $L$ such terms, so from (1), you only find that the energy is $O(\epsilon L)=O(1)$, which comes as no surprise.
So to make sure this energy is actually vanishing as $L\to\infty$, we have to make sure that in fact $\epsilon\propto 1/L^2$ or smaller. How can this be the case?
Clearly, this requires that the first-order term in (1) when expanding
$$
U_{g^{-1}h} = 1+\epsilon G+O(\epsilon)
$$
vanishes; this term is nothing but
$$
R:=\epsilon\,\mathrm{tr}[h(I\otimes G)\rho-h\rho(I\otimes G)]\ .
$$
It is easy to see that $R\stackrel{!}{=}0$ does not follow from the symmetry of the Hamiltonian. Instead, it must be a property of the ground state, or the ground state and Hamiltonian together!
Indeed, imagine $\Delta e = \epsilon K + O(\epsilon^2)$ for some non-zero constant $K$. Then, summed up over the whole system, the total energy difference is
$$
\Delta E = \epsilon L K + O(L\epsilon^2)\ .
$$
Indeed, if $\epsilon>0$, this gives a term or order $1$. However, we can also change the sign of $\epsilon$, such that $\epsilon L K<0$ -- in that case, we get an ansatz for a state such that
$$
\Delta E < 0\ !
$$
Clearly, this is impossible since $\Delta E$ is the energy difference to the ground state!
Thus, the fact that we are in a ground state rules out a term linear in $\epsilon$, and thus, the energy per site/boundary is $O(\epsilon^2)=O(1/L^2)$, and thus the total energy difference is
$$
\Delta E = O(1/L^2)\ ,
$$
where $L$ can be taken to be the system size (if we twist at each site by an angle proportional to the position).