Is a Banach space also a metric space? Since a Banach space is a complete normed vector space and a norm always induces a metric, a Banach space must be a metric space, right?
If so, why is a Banach space defined as a complete normed space and not a complete metric space?
 A: It was already covered in comments, but a complete normed vector space has a lot more structure than just a complete metric space. Every normed vector space is a metric space but not the converse.
With a normed vector space, you have to have some idea of adding the elements, a field of scalars, an idea of scalar multiplication. In arbitrary metric spaces you don't have these.
A: A complete metric space need not be a complete normed space. A complete normed space is also called a Banach space! Since every norm induces a metric, these Banach spaces reside in the collection of all complete metric spaces. Thus, complete metric space and complete normed space - two different notions but related indeed.
A: Every Banach space is a metric space. However, there are metrics that aren't induced by norms. If this were the case, then Banach spaces and complete metric spaces would the same thing... if metric spaces had operations and an underlying field! The difference is more than just the metric/norm dichotomy. For completeness of the answer (no pun intended), a metric $d:X\times X \to \Bbb R_{\geq 0}$ comes from a norm if and only if: $$d(x+z,y+z) = d(x,y)\quad\text{and}\quad d(\lambda x,\lambda y) = |\lambda|d(x,y),$$ for all $x,y,z \in X$, for all $\lambda \in \Bbb R$. If this is the case, the norm is $\|x\| = d(x,0)$.
