Why does the breaking of a 'continuous' symmetry lead to gapless excitations but not that of a discrete symmetry? Goldstone theorem, especially in the context of condensed matter physics, can be stated as:

Whenever there is spontaneous breakdown of a continuous global symmetry, the spectrum of the theory contains gapless excitations. However, this is not true for breakdown of discrete symmetries.

For example, the spin wave of Heisenberg ferromagnet is a classic example of a gapless excitation resulting from the spontaneous breakdown of the spin rotational symmetry of the Heisenberg model. There exists rigorous, formal proofs of this theorem.
Question But physically, can we understand why this happens (or doesn't happen)?
It's true that in the Heisenberg model $$H=-J\sum_{\langle i,j\rangle}{\bf s}_i\cdot {\bf s}_j\tag{1}$$ the spins ${\bf s}_i$ can rotate in ${\rm 3D}$ while the Ising spins can only point up or down. It's tempting to think, if the spins can rotate continuously, we can excite the system with arbitrarily small energy by gradually tilting the spins infinitesimally w.r.t its immediate neighbors. This is not allowed if the spins are only allowed to flip (e.g., Ising model) and therefore, it requires a finite energy cost to excite the system! In the first case, we thus expect gapless excitation and gapped in the second. But this argument (based only on the freedom of movement of spins) is half-baked!
Continuous movement in ${\rm 3D}$ doesn't seem to ensure emergence of gapless excitations. The Hamiltonian must also have a continuous symmetry. For example, if we think of a generalized Heisenberg model $$H=-\sum_{<ij>}(Js_{ix}s_{jx}+J's_{iy}s_{jy}+J''s_{iz}s_{jz}),\tag{2}$$ the spins can still rotate continuously in ${\rm 3D}$, but the Hamiltonian of Eq.$(2)$ has no rotational symmetry unless the constants $J,J',J''$ are all equal (in which case $(2)$ reduces to form $(1)$)! Therefore, Goldstone's theorem cannot be applied here and there is no certainty that the spectrum will have gapless excitations.
So the question is why is it necessary to have a continuous symmetry in the first place to have gapless excitations? I am not looking for a formal proof.
 A: DISCLAIMER: the heuristic argument provided below largely pertains to classical systems. I make no pretense of rigor, nor of simple generalization to quantum systems.

It's tempting to think, if the spins can rotate continuously, we can excite the system with arbitrarily small energy by gradually tilting the spins infinitesimally w.r.t its immediate neighbors.

Let me start by reviewing the basic physical argument for Goldstone modes. If a continuous global symmetry is broken in the ground state, then the ground state picks out one particular broken symmetry state. By slowly varying the system over space by the continuous symmetry, the system always appears to be in the ground state locally. Globally, there can be as small a "twist" as you want, so the energy must be gapless.
Now, suppose your system has continuous degrees of freedom, but does not have a continuous symmetry. Then by slowly varying the spins, there is no way the system could appear to be locally in the ground state everywhere -- indeed, if it did look like this, then you would certainly have a continuous symmetry! Instead, your system will have larger and larger energy densities the more you "twist".
A: Short answer: There are other gapless excitations than Nambu-Goldstone modes.
Longer answer:
The Goldstone theorem guarantees that there is at least one gapless excitation if a continuous symmetry is broken spontaneously, in the absence of long-range forces. That's it. There could be plenty of other gapped and gapless excitations other than the Nambu-Goldstone mode(s).
In fact, the theorem it is constructive, and will tell you how to define this excitation. Namely, by acting with the broken symmetry generator on the broken-symmetry state. Therefore, this excitation is sometimes called a "non-linear realization of the symmetry".
A: 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).
A: 
But this argument (based only on the freedom of movement of spins) is half-baked! Continuous movement in 3D doesn't seem to ensure emergence of gapless excitations.

No, certainly not, but it's close. You're skipping over the part about the energy cost.  In your proposed generalized Heisenberg model, $J\neq J'\neq J''$ would generically imply the existence of one or more discrete ground states, separated from one another by higher energy configurations.  Starting from the ground state, it would require a finite amount of energy to excite the system, which means that it's gapped.
If all of the $J$'s are very close to one another, then you have an approximate (not exact) symmetry, which would correspond to low-energy (but not exactly massless) excitations.  This is the case e.g. for pions, which can be thought of as the pseudo-Goldstone modes corresponding to the approximate chiral-flavor symmetries which are broken by the different masses of the quarks, and which therefore have far smaller masses than most hadronic particles.
So, it's not continuous movement but rather continuous movement at no energy cost which implies the existence of massless Goldstone modes. If the energy cost is small, then the corresponding pseudo-Goldstone modes will be light, and we would say that we have an approximate (or weakly-broken) symmetry.
