s: Subshell of an atom has spherical shape.
p: Subshell is a dumb-bell shaped with a node at center of the atom.
If we put two shells together they would intersect in an atom. In such a model, electrons of the s subshell will collide with the p subshell. This will lead to instability of the atom and its model. Is this correct? Where am I wrong?
I will try to address the points you have made in your question, one by one.
First, what are electrons?
Thanks to @CuriousOne, electrons are best treated as perturbations in a quantum field, which is explained by quantum-electrodynamics. In the Standard Model, they are considered elementary particles of the 1st generation lepton family with a mass 1/1836 times that of a proton and 1 unit negative elementary charge. It is in fact what defines elementary electronic charge.
Firstly, how do electrons stay bound inside an atom?
Short answer: Not necessarily.
Long answer: An electron is allowed infinitely many paths about an atom, even if it means going around the universe. However, many researchers contributed towards finding a solution to this infinite-histories/infinite-paths problem, most prominently Prof. Richard Feynman. (I suggest you read his book "The Feynman Lectures on Physics", for a good introduction to quantum mechanics, among other things.).
It was calculated that the electron's probability was highest for being limited to an atom, and that too in some certain limited geometrical distributions for very simple atoms (one-electrons systems or hydrogen-like species). These limited geometrical distributions in which the probability of finding an electron is highest are called "orbitals". The probability of finding an electron at a particular coordinate is obtained by squaring the value of the electron's Schrödinger wavefunction at those coordinates.
For your reference, here is the Schrödinger equation in one dimension:
$$i\hbar\frac{\partial \Psi}{\partial t}= -\frac{\hbar^2}{2m}\frac{\partial^2 \Psi}{\partial x^2} + V(x)\Psi(x,t)\equiv \tilde{H}\Psi(x,t),$$
where $i$ is the imaginary unit, $\Psi$ is the time-dependent wavefunction, $\hbar$ is the reduced Planck's constant, $h/2\pi$, $V(x)$ is the potential, and $\hat H$ is the Hamiltonian operator.
Also, here are the probability plots for the first three orbitals of hydrogen-like atoms:
And all of them in a hydrogen-like atom:
The solution for the Schrödinger equation is very complex and at best approximate for multi-electron atoms, and that too using spherical rather than Cartesian coordinates which represent the electron-waves as standing waves in spherical harmonics (it is like a standing wave on a circular string, except this is like a standing wave throughout some volume of an orbital region) (thanks @user36790). There are inter-electronic repulsions (nuclear screening) which needs to be accounted for, which greatly increases the complexity for even a 2-electron system (helium-like species).
Finally, what you asked in your question: Why don't electrons in overlapping orbitals collide?
Answer: They might, but it makes little to no sense to say that two waves "collide" with each other. They may superpose or interfere, but not "collide" in the classical sense.
Overlapping of adjacent orbitals leads to orbital degeneracy and reorganization of orbitals (although insofar as much hybrid orbitals are mathematical manipulations) (remember the chemistry of complexes and degenerate d-orbitals of transition elements?), among other things. Orbital boundaries are not sharply defined. They, in practice, nearly always overlap to some extent, but their electrons remain distinct.
However, two electrons cannot in-fact collide with each other, as in such a condition the wavefunction becomes undefined as two electrons are occupying the same state momentarily, but thankfully the Pauli exclusion principle prevents exactly this. The Pauli exclusion principle expressly states that two electrons in the same atom cannot have the same values for all of their quantum numbers (n, l, m and S) for any period of time. So you can be rest assured that two electrons in overlapping orbitals do not and can not "collide", even if for no other reason than their mutual electrostatic repulsion.
The question is within a Bohr model of an atom, and the Bohr model worked well for the hydrogen atom by postulating fixed orbits, but it is not the real description of what happens at the atomic dimensions. The real description for the hydrogen atom is given by the solutions for the hydrogen potential of the Schrödinger equation with the postulate of the Born rule. This gives rise to the orbitals for the electron in the hydrogen atom.
Orbitals are a probability locus/position they answer the question: what is the probability if I do an experiment that the electron will be found in an $( r, \theta, \phi )$ spot at time t. This is simple for the hydrogen atom since there is one electron.
For more complicated atoms the quantum mechanical solutions are complicated, but still there are orbitals, and the shapes of S and P are as you describe. There is another overall rule though that keeps one electron in each available quantum orbital and no overlaps except in the visualizations. It is the Pauli exclusion principle which keeps one electron at a time at each orbital, and the electrons not interacting.
Actually, as one can see from the orbitals, there is a probability of the electron to be found within the nucleus of the atom, and there, there is no Pauli exclusion for the electron, and in a suitable nucleus this has been observed, called electron capture.
There are many ways... Due to their variable oxidation state, due to the strong nuclear force, and poor shielding effect.
Electrons don't collide (unless they are subjected to very high velocity). They usually repel. But this repulsion force can be cancelled if the nuclear force is too high. Hence they be together.
In the case of variable oxidation state, the s and p orbitals have very similar energy (very less energy difference). Hence they form pairs and become stable.
