If like charges repel, why doesn't a charge break itself apart?
Composite particles like protons don't break apart because of the strong interaction which holds their constituents (the quarks) together. Elementary particles like electrons don't break apart because they are point-like particles, i.e. they are not made of “parts” (if the Standard Model is correct).
This was one of those big questions in the 19th century. It still causes some consternation. If you have a composite system, such as the nucleus of an atom, some other force is necessary. This force of course is the nuclear interaction. This keeps the protons from flying apart, though for some unstable nuclei there are transitions that eject charged particles, electrons or positrons, due to weak interactions. In the case of the proton it is composed of three quarks and these are bound to each other by the QCD (quantum chromodynamics) interaction. The gauge bosons called gluons interact most strongly at low energy and these keep the quarks, with charges $2/3,2/3,-1/3$ in a bound state.
Things are a bit more mysterious with point-like particles, such as the electron and other leptons and quarks. We generally do not regard such particles as composite, though this has not stopped people from proposing constituents called preons or rishons that make them up.
There is a problem with defining the mass of the electron or any point-like electrically charged particle. The mass of the electric field is $$ m_\textrm{em}~=~\frac{1}{2}\int E^2~\mathrm d^3r~=~\frac{1}{2}\int_r^\infty\left(\frac{e}{4\pi r^2}\right)^24\pi r^2~\mathrm dr~=~\frac{e^2}{8\pi r}. $$ if the electron has zero radius this is divergent. There is the classical radius of the electron $r~=~\alpha\lambda_c$ $=~2.8\times10^{-13}~\mathrm{cm}$ for $\lambda_c~=~\hbar/mc$ the Compton wavelength. This raises some questions, for the classical radius suggests "structure," and it also has a relationship to something called Zitterbewegung.
A more standard approach to this is renormalization. A screenshot of this is to look at this integral with the variables $p~=~1/r$ so in this integral above $\mathrm dr/r~\rightarrow~-\mathrm dp/p$. Here we are thinking of momentum and wavelength or position as reciprocally related. This integral is then evaluated for a finite $r$ as equivalent to being evaluated for a finite momentum cut off $\Lambda$ $$ I(\Lambda)~=~\int_0^\Lambda\frac{\mathrm dp}{p}~\simeq~1~+~2^{-1}~+~3^{-1}~\dots $$ which is equal to $$ \lim_{\Lambda\rightarrow\infty}I(\Lambda)~=~-\zeta(1) $$ In some ways this is a removal of infinities. Another curious way to look at this is with $p$-adic number theory. This is a topic that could consume a lot of bandwidth.
We have another way to look at this. This comes down to the question of what do we mean by "composite." It also forces us to think about what we mean by the locality of field operators. The Dirac magnetic monopole is a solenoid with an opening to an infinite coil. The condition for the Dirac monopole is that the Aharonov-Bohm phase of a quantum system is zero as it passes the "tube" of the solenoid $\psi~\rightarrow~\exp\left(ie/\hbar\displaystyle\oint{\vec A}\cdot ~\mathrm d{\vec r}\right)\psi$. This might be compared to "cutting off the tail" on the magnetic monopole charge. The vanishing of this is equivalent to saying $$ 2\pi N~=~\frac{e}{\hbar}\displaystyle\oint{\vec A}\cdot ~\mathrm d{\vec r}~=~\frac{e}{\hbar}\iint\nabla\times{\vec A}\cdot{\vec a}, $$ for the integral evaluated over units of area of the opening. This is of course the magnetic field ${\vec B}~=~-\nabla\times{\vec A}$ evaluated in a Gauss' law that gives the magnetic monopole charge $g~=~\displaystyle\iint\nabla\times{\vec A}\cdot{\vec a}$ and we use this expression to see the S-duality relationship between the electric and magnetic monopole charge $$ eg~=~2\pi N\hbar, $$ sometimes called the Montonen-Olive relationship.
This means that if we have an electric charge we can use the renormalization machinery to illustrate how the vacuum around it is polarized with virtual particles according to $\alpha~=~\frac{e^2}{4\pi\epsilon\hbar c}$. The electric charge is comparatively weak in strength with a modest polarization of the vacuum expanded in orders of $\alpha$ for $N$ internal lines or loops. This S-dual relationship tells us that while this is modest, the magnetic monopole is very strong and the vacuum is a "bee's nest" of lots of particles. This then means the dual of the electric field is a magnetic monopole field that in some ways appears composite.
This means in some ways we have questions needed to be asked about the locality of field operators. Something that appears local, point-like and "nice" may be dual to something that appears not so local, more composite-like and not renormalizable. As a result there are still open questions on this, and even Feynman agreed with Dirac that the situation with QED was not perfectly satisfactory.
When you have a charged object, for example a charged metal sphere, of course the charges on the surface of the sphere interfere with each other. Because of these effect the charge gets distributed equally over the sphere.
However, these effects are not big enough to actually break up the sphere or something like that.
If your object is charged high enough there can be discharges to other objects (like the air) because of the potential difference.
Looking at an electron: It is a subatomic particle and can not be split in "half". It is not like an object that carries the charge. The charge is a fundamental property of the electron itself.
Protons contain two $\frac{2}{3}$-positive charged up-quarks and one $\frac{1}{3}$-negative charged down-quark. The quarks are glued together through gluons. These bring up the necessary force to keep the proton together. Quarks are also elementary particles, as per the Standard Model
Using a particle accelerator you can smash protons and other particles into each other. In this case it is possible to destroy the bindings between the quarks and new particles are created.