Confusion about security of quantum key distribution In Quantum Key distribution, it is my understanding that when Eve tries to intercept photons being sent from Alice to Bob, she has a 50% chance of using the correct polarization filter, and so 50% chance of getting the correct bit information. Let's just say hypothetically that she somehow gets all the photons she intercepted using the correct filter (and hence correct bit information), and re emits them so that they're all exactly the same photons as Alice had intended them to be. 

Doesn't this mean that during the error reconciliation between Alice and Bob, they'll detect no errors and hence conclude that the distribution of the key was secure, but in reality, Eve has perfect information about the "secret" key? 
Doesn't this violate the proof that QKD is "provably secure"?
You are right and you are wrong. Indeed, in the scenario that you describe Eve would perfectly break the system (in a certain way). She would learn information about the secret key without being caught. Note, however, that this only has any chance of success for a small amount of bits. The chance of success when measuring $x$ qubits is $2^{-x}$ (in this idealized setting ignoring all the implementation attacks). That's why you cannot use the the plain exchanged bit string. Instead, you have to use privacy amplification. There are different methods for this. The main idea is that you use some function to turn a long bit string of which Eve possibly knows some bits into a short bit string about which Eve has no information as long as she did not know a lot of the input bits. Choosing parameters right, you can that way ensure that as long as Eve does not learn more than e.g. $128$ bits of the original bit string the output string (used as secret key later) has e.g. $128$ bits of entropy.