According to the book I'm reading, the punctured disk is neither open nor closed. 
I understand why it isn't open, namely because it contains boundary points. But why is it not a closed set? All the points |z| = 1 or the boundary points are contained in the set.
Unless they consider the point |z| = 0 to be a boundary point. Then the conclusion to why it is not closed is that the set does not contain the point|z| = 0.
Is my reasoning correct or is there some other reason I'm not seeing as to why it would not be closed.
Also, in this context, just to clear up confusion, the set of points that make up the unit circle |z| = 1 is a closed set, right? Even though it doesn't contain points |z| < 1. I'm unsure about the meaning of the closure of a set and a closed set. Thank you.
A set is closed if its complement is open.
The complement of the punctured disk consists of a neighbourhood of $\infty$, which is open, and a single point $\{0\}$.
By definition of open sets, if $\{0\}$ was open, $\forall \epsilon > 0, \exists z \neq 0, |z - 0| < \epsilon \land z \in \{0\}$, but that is impossible. Hence the complement of the punctured disk is not open.