Is a vector space a subspace of itself? 
We know that a subspace (of a vector space $V$) is a vector space that follows the same addition and multiplication rules as $V$, but is a vector space a subspace of itself?

Also, I'm getting confused doing the practice questions, on when we prove that something is a vector space by using the subspace test and when we prove V1 - V10, which are the ten axioms of vector spaces. So for example in $\Bbb R^2$, we have that $\vec{x} + \vec{y} = \vec{y} + \vec {x}$, etc..
 A: I'm guessing that V1 - V10 are the axioms for proving vector spaces.
To prove something is a vector space, independent of any other vector spaces you know of, you are required to prove all of the axioms in the definition. Not all operations that call themselves $+$ are worthy addition operations; just because you denote it $+$ does not mean it is (for example) associative, or has an additive identity.
There is a lot to prove, because there's a lot to gain. Vector spaces have a simply enormous amount of structure, and that structure gives us a really rich theory and powerful tools. If you have an object that you wish to understand better, and you can show it is a vector space (or at least, related to a vector space), then you'll instantly have some serious mathematical firepower at your fingertips.
Subspaces give us a shortcut to proving a vector space. If you have a subset of a known vector space, then you can prove just $3$ properties, rather than $10$. We can skip a lot of the steps because somebody has already done them previously when showing the larger vector space is indeed a vector space. You don't need to show, for example, $v + w = w + v$ for all $v, w$ in your subset, because we already know this is true for all vectors in the larger vector space.
I'm writing this, not as a direct answer to your question (which Jose Carlos Santos has answered already), but because confusion like this often stems from some sloppiness on the above point. I've seen many students (and, lamentably, several instructors) fail to grasp that showing the subspace conditions on a set that is not clearly a subset of a known vector space does not prove a vector space. The shortcut works because somebody has already established most of the axioms beforehand, but if this is not true, then the argument is a fallacy.
You can absolutely apply the subspace conditions on the whole of a vector space provided you've proven it's a vector space already with axioms V1 - V10.
A: Yes, every vector space is a vector subspace of itself, since it is a non-empty subset of itself which is closed with respect to addition and with respect to product by scalars.
