In what way do Cooper pairs of electrons bond and stay bonded in superconductors? I understand how electrons initially move into another's vicinity, but nowhere can I find a fathomable answer to this. Also, does the pairs forming 'a condensate' mean a Bose-Einstein condensate?
 A: They don't really bind, the interaction can be too weak to bind, but the pairs condense anyway. The theory is interesting precisely because the condensation is happening from an interaction too weak to form actual pairs. When the interaction is strong, the pairs can be considered as effective particles, and the theory of superconductivity is different and simpler--- it's a BEC of the pairs.
The standard description is BCS theory, and I'll explain it Bogoliubov's way, which is found in countless places. I'll consider the electron field interacting with a nonrelativistic instantaneous potential. This is not accurate in traditional superconductors already, because the interaction is phononic and retarded, but whatever, the phenomenon doesn't care about this very much.
You have an electron field $\psi(k)$, and the interaction Hamiltonian is
$$ V(k_\mathrm{in}-k_\mathrm{out}) \sum_{s} \bar{\psi}^{s_1}_{k_1}\bar{\psi}^{s_2}_{k_2} \psi^{s_3}_{k_3} \psi^{s_4}_{k_4} \delta(\sum k), $$
where the $\delta$ is multiplied by 3 $2\pi$ factors as always, I assume V is spherical symmetric, and there are two spin fluids which are degenerate in energy. The main assumption, which is justified self-consistently, is that the field $\psi^0_{k}\psi^1_{-k}$ has an expectation value. You can see this is self consistent from the fact that if V is negative at $2|k_f|$, this expected value lowers the energy classically.
The expected value of $\psi\psi$ is a type of condensation, but it isn't a local condensation because there is no real local bilinear field that it is the exectation value of. In the limit that the coupling is strong, you can make a local field which creates a bound Cooper pair, but this is only linked adiabatically to the BCS picture.
A: Yes, the cooper pairs form a BEC. See Kardar's Statistical Physics of Particles for a detailed derivation of this. I think Kittel's Thermal Physics might also have some explanation (on an easier level).
For a BEC to form, the bosons should be non-interacting. To good approximation, this is the case for cooper pairs.
A: The attraction is caused by screening, which means that when two electrons come together, they tend to push away the nearby electrons, which leaves a net positive charge in the area that keeps those two electrons together.
