Cantor space: example of an open set that's not closed?

Hint: Look at the complement of a point—the Cantor space is not discrete.

Later: Since the Cantor set is homeomorphic to a countable product $\prod_{n=1}^{\infty} \mathbb{Z}/2\mathbb{Z}$ of the cyclic group of order two (use the identification of the cantor set with the points in $[0,1]$ whose infinite ternary expansion contains no $1$), it is homogeneous. This means in particular that the Cantor set has no isolated points and hence it has no open points.

Now note that a basic open set is of the form $\prod_{n=1}^{\infty} X_n$ with all but finitely many $X_n = \mathbb{Z}/2\mathbb{Z}$. But this means that a basic open set contains a space homeomorphic to the entire Cantor space, hence non-empty open sets are uncountable (in fact of cardinality $\mathfrak{c}$). In particular we see that a convergent sequence (which is of course closed) can't be open. Passing to complements we get an open set that's not closed, as required.

[Meta: Thanks to ccc for pointing this out and to Brian for making me think again.]

Just make a sort of fishbone, and remove the spine: $$\bigcup_{i=1}^{\infty} [10^i].$$