Do subspaces obey the axioms for a vector space? If I have a subspace $W$, then would elements in $W$ obey the 8 axioms for a vector space $V$ such as:
$u + (-u) = 0$ ; where $u ∈ V$ 
 A: Yes, indeed. A subspace of a vector space is also a vector space, restricted to the operations of the vector space of which it is a subspace. And as such, the axioms for a vector space are all be satisfied by a subspace of a vector space.
As stated in my comment below: In fact, a subset of a vector space is a subspace if and only if it satisfies the axioms of a vector space.
A: This property should normally be the definition of subspace. That is, if foobar denotes a kind of algebraic structure, then the generic definition of a sub-foobar $A$ of a foobar $B$ is that $A$ is a foobar with the properties that the underlying set of $A$ is a subset of $B$ and all operations on $A$ are in fact the restrictions to $A$ of the corresponding operations on $B$.
However, in order to test that a given subset of $B$ is a sub-foobar (with the restricted operations) it is typically not necessary to verify all foobar-axioms because some follow automatically (e.g. associativity of operations). Hence, there are often simplified sub-foobar criteria. (For eaxmple, if foobar=finite group, "nonempty and closed under the operation" is enough to test for a subgroup).
If a text uses such a simplified criterion as the definition of subfoobar (and shows that a sub-foobar is a foobar two pages later), I think this is bad didactics.
A: Yes, exactly: a subspace is also a vector space with respect to the (restrictions of the) same operations.
A: Yes. Note that one of the ways to check its a subspace is to verify all of the axioms and then show that $W$ contained in $V$, while the other method would be to use the subspace test which checks closure under addition and scalar multiplication (and nonemptyness).
