I don't get band structure of solids If the energy levels of bound electrons are discrete, why do band structures in solids arise?
 A: Ok, I am by no means and expert on solid state but I might be a little helpful.
Band structure in solids arises due only to periodicity of the lattice. It all comes down to this periodicity.
Periodicity of the lattice makes the potential also periodic.
This periodicity has many (interesting) consequences (Bloch states and bla bla bla) but the one that matters to you is that energy eigenstates of the Hamiltonian are specified by $\vec{k}$, a vector of the reciprocal space, and another index that I will call "n" (this "n" will label the bands).
Now, the reciprocal is periodic becuase the crystal is periodic. Then 
$\epsilon_n(\vec{k})=\epsilon_n(\vec{k}+period)$
So you see that this energy spectrum of the single electron in the periodic potential of the crystal has to be bounded and all energy values for a given $n$ must be in a band of possible values. THIS is the band structure. You have a band for every "n".
Now, are all $\epsilon_n(\vec{k})$ allowed? No, (at least no if you work under Born and Von Karman Boundary conditions( which is what I have always seen done)). This boundary conditions discretize the posible $\vec{k}$ vectors (and actually you can limit yoursef to the $\vec{k}$ of a primitive cell of the reciprocal lattice because of Born and Von Karman).  
So you see, electrons in crystals are also in a discrete spectrum.
This is quite nicely explained in Ashcroft's book. The idea is, periodicity of the lattice gives you band structure of the energy levels of the single electron. Then Born and Von Karman boundary conditions discretize the spectrum.
A: When solving for the spectrum, you assume the solid has infinite extent, and so the electron is not really bound. Thus you can get a continuum of momentum. 
Consider for example the trivial case where there are no atoms, and you just have a free electron in space. Then the energy is proportional to momentum squared, and momentum can be arbitrarily big, so you just get one big band.
The bands come from the interaction of the electron with the lattice. Suppose we can add in a lattice of atoms with which the electron interacts. How does this change and cause us to have several distinct bands? Well notice the potential from the atoms breaks continuous translational invariance (and therefore momentum conservation). However, since the atoms still lie on a lattice we have invariance under translations by a lattice vectors, and so we have a weaker form of momentum conservation, which says that wavenumber is conserved only up to a wave-vector of the reciprocal lattice to the lattice our atoms are on. 
But our energy spectrum still seems connected, how do we get band gaps from non-conservation of momentum? Now consider a reciprocal lattice wave-vector $\vec{K}$ and electron wave number $\vec{k}$ near $\vec{K}/2$. Then symmetries laws do not forbid a transition from wave vector $\vec{k}$ and wavevector $\vec{K}-\vec{k} = \vec{K}/2 + (\vec{K}/2 - \vec{k}) \approx \vec{K}/2\approx \vec{k} $ since $\vec{k} \approx \vec{K}/2$. Now since $\vec{K}-\vec{k} \approx \vec{k}$, the states corresponding to these two wavenumbers have similar $k^2$, i.e., similar unperturbed energy. But now with the lattice in place, there will generically be some matrix element allowing transitions between these states and so these energy levels will split even though they were originally close. Energies near $\dfrac{(\vec{K}/2)^2}{2m}$ therefore are not attained because of this splitting and so there is a gap in the spectrum. This is the band gap.
A: Because in solid states, the electrons are not bound.
A simple derivation of the continuous bands can be made if you look at particle in a box (Wikipedia).
If we approach the number of an infinite box with infinite particles (and constant density), there are infinite states, and the states get closer, as the energy gap goes as $L^{-2}$.
Therefore a continuous band arises.
The derivation of the band gaps is a bit more complicated.
In the simple one dimensional case, this is related to the periodicity of the atoms. Looking at the wavelength, that is the same as the periodicity of the crystal, there are basically two options: the one where the electrons have a high density probability at the atoms, therefore they have a lower energy, as the are attracted. The other possibility is that they have a higher probability between the atoms. This is energetically not so good for them. So we see we have 2 states with the same $k$-vector, but with different energies. This is the reason for the band-gap.
A: You could think of making a solid by slowly bringing all the atoms closer together.  As they get closer the discrete electron orbits start to bump into each other.  But the electrons can't be in the same state, and you start to get new states that are a combination of the discrete states.. And these form bands.(Yeah that last is a bit of a cop out, see a solid state text.)  Note: you have bands in insulators where the electrons are not free.
Weird, what I'm pretty sure is the right answer gets voted to the bottom.  If I edit this will it get bumped up to the front of the line again?  So putting atoms in a periodic array does not guarantee that bands will form.  When atoms get close to each other the electrons are forced into bands because they are fermions and can't be in the same state.  
