This set is not closed there are non-rational numbers in that interval which are limit points of sequences that consist only of rationals. For example, any of the algebraic numbers. I think that all real numbers are limits of such sequences, but I might be mis-remembering some subtlety of Dedekind Cuts (one method for constructing the Reals).

This set is not open because any rational is the limit of a sequence of non-rational reals. This probably makes intuitive sense, but just for the sake of formality: To construct such a sequence for any rational r, start with the number x_1 = 1/pi, and approach by a factor of 1/pi at each step, i.e. x_n+1 = x_n + (r-x_n)/pi . x_n is irrational because pi is transcendental.

Any simple interval in R will be either closed or open on each end (but it could be closed on one and open on the other). It's more illustrative to create a set with a non-compact interior. In higher dimensions it's possible to have more exotic borders on an interval, but I think that border will just end up being isomorphic to a non-compact set in a lower dimension.

Proof that it's not closed: the sequence (1, 1/2, 1/4, 1/8, ...) is entirely inside the interval, but converges on 0, which is outside the interval, therefore etc.

Proof that it's not open: the sequence (2, 3/2, 5/4, 9/8, ...) is entirely outside the interval (i.e. inside the complement), but it converges on 1, which is inside the interval (i.e. outside the complement). Thus the complement of the interval is not closed, therefore etc.

http://en.m.wikipedia.org/wiki/Clopen_set