Well for part a) you got the right answer, but I am not sure for the right reason. You showed the set is in fact not open because I think what you meant to point out for instance there is no open nbhd about (1,0) which is contained in S, so it is not open. But you must make an argument for why it is not closed too. If it were closed, its complement would be open. But its complement has the origin as an isolated point. There is clearly no open nbhd about the origin which would be contained in the complement of S, so the complement is not open thus S is not closed.

Your interior is correct.

2)

This is just basically a square centered at the origin with side length 4.

That is . This set is closed because it's complement is open. Think you can prove this? Pick an arbitrary point outside of the box and show there is an open nbhd contained in the complement of the box. The interior will be See why?