Your existing answer talks about quark confinement, but stable nuclei can't really be described using quark and gluon degrees of freedom. Also your existing answer doesn't answer your title question: why don't nuclei collapse to a point?
To first approximation, nuclei do collapse into a point. The diameter of a nucleus is typically about $10^{-5}$ the diameter of an atom, which means the nucleus occupies something like $10^{-15}$ of the atomic volume. If your atom were the size of a house, the nucleus, to scale, would be the size of a grain of salt (and yet containing 99.95% of the atom's mass).
In nuclear physics it makes more sense to talk about energy than it does to talk about force. The two concepts are closely related. Two protons separated by a distance $r$ have an "electrical" interaction energy
$$
U_E = +\alpha\frac{\hbar c}{r}.
$$
Here $\hbar c = 200\rm\,MeV\,fm$ relates distance and energy.
The fine structure constant, $\alpha \approx 0.0073 \approx 1/137$, represents the strength of the electrical interaction.
The positive sign, and the $r$ on the bottom, mean that bringing two protons closer together stores energy in the electric field between them.
If you find yourself with two protons at rest some distance apart, they'll tend to evolve to reduce the energy stored in the electric field by moving yet further apart; this is the sense in which the electric interaction is repulsive.
The electric field isn't the only place where interacting protons can store energy. There is also the pion field, whose interaction energy is given by a Yukawa potential,
$$
U_\pi = -\alpha_\pi \frac{\hbar c}r \times e^{-m_\pi r}.
$$
This is mostly the same as the electrical potential, but the differences are important:
The sign is negative: protons liberate energy from the pion field by approaching each other, so the force is attractive.
There's a different coupling constant, $\alpha_\pi$. If I'm converting units correctly, the pion coupling constant is something like $\alpha_\pi\sim100$: the energy associated with the "pion exchange" interaction may be as much as ten thousand times stronger than the energy associated with a "photon exchange" interaction.
Critically, there's an exponential factor that depends on the mass of the pion, $m_\pi$. To be proper this should have some factors of $\hbar$ and $c$ to make the argument of the exponential dimensionless; alternatively we can be cleverer about units and not have any distracting cruft. This means there is a length scale $r_0 \propto 1/m_\pi$ beyond which the pion interaction totally dies away, but for very short distances the interaction looks like electricity with a different coupling constant.
These differences address a little of your confusion. The strong interaction is much stronger than electromagnetism, but not infinitely stronger. But your title question still remains: why doesn't the nucleus collapse exactly to a point? That's because we still haven't exhausted the forest of different ways that two protons can store energy. The next two that are important are the rho and omega fields,
\begin{align}
U_\rho &= +\alpha_\rho \frac{\hbar c}r \times e^{-m_\rho r},
\\
U_\omega &= +\alpha_\omega \frac{\hbar c}r \times e^{-m_\omega r}.
\end{align}
Like the pion field, these meson interactions "turn off" when the nuclei separate from each other, but the distances are shorter than for the pion because the masses are larger. So the interaction between two protons has at least three different regions:
- Distant protons are repelled by the electrical interaction.
- Very close-together protons are repelled by vector meson interactions.
- At intermediate distances, protons can be attracted to each other by pion interactions.
Here's a plot using the potentials we've discussed so far:
This simple model reproduces several real features of real nuclei:
- In uniform nuclear matter, nucleons are separated by about 1.2 fm.
- The energy required to remove one nucleon from a nucleus is typically about 10 MeV.
- In very heavy nuclei, the length scale for nuclear attraction is shorter than the diameter of the nucleus; such nuclei are unstable.
You can't tell from this plot, but the total interaction energy becomes positive --- that is, electrostatic repulsion wins out over strong attraction --- at a separation of about 12.5 fm.
Uranium-235, famously unstable, has diameter 14 fm.
Uranium is unstable because protons at one end of the nucleus are repelled by, rather than attracted to, protons at the other end of the nucleus.