Proving that subspaces of space of all polynomials $\mathcal{P}$ is a vector space I'm trying to write a proof, that some certain spaces of all polynomials are vectors spaces. If we take field $\mathcal{F} = \mathbb{C}$ and $V = \mathcal{P}$, then if I understand correctly, I have to prove all axioms of vector space, because it is not a subset of some more widely known vector space (as example if we took space based on $\mathbb{R}^2$).
So, will it be a generally good tactics to first prove that general $(\mathcal{P}, \mathbb{C}, \oplus, \odot)$ is a vector space, and then state that some spaces, which were given to me are just subspaces of it, requiring to prove just existance of $\vec{0}, -\vec{x}$ to be a vector space?
 A: The space of all polynomials over a given field (with the obvious addition and scalar multiplication operations) is widely known to be a vector space over that field.
I'll leave it to you to judge whether it's widely known to your audience (or, if you are a student, whether your instructor trusts it's known by you or you can convince your instructor of such).
A: You would have to prove all the axioms. Then afterwards, given a subset of it, you'd only have to check for the 0 element, and closure of addition and scalar multiplication. 
A: Indeed, it is much easier to show that something is a subspace than to independently verify all the axioms for a vector space. For example, let $F$ be a field, and take it's polynomial ring $F[x]$. One way to think about this is the fact that it is an abelian group, and that scalar multiplication agrees with its group structure. Now if you wanted to show that
$$\{P(x) : \operatorname{deg}P \leq 2\}$$
is a vector space, showing that it is a subspace is in general much easier than doing all of the annoying work again. 
