you can prove why
b) Taken an open cover of call it A. Look at B where B consists of the inverse of the projection map of each open set in A. For instance if then the corresponding open set in B is . This is an open set since it is a product of two open sets in , so we notice B is an open cover of K, a compact set, so it has a finite subcover. Then look at the projections of each of these sets and it is a finite subcover of and thus its compact.
I dunno how to do the fancy script A, so I apologize for the awkward notation of a representing a set and A being a collection of sets...
c) This is the diagonal and it is probably a theorem you have done that is pretty easy to show from definitions that hausdorff iff diagonal is closed so is closed in but its projection is which is open.
If you dont believe me that is closed, just look at and see why it's open.
Hope that helps.