$V$ has a basis $B \subset V$, s. t. every $v \in V$ can be written as a ﬁnite sum? I'm reading "Sets, Models and Proofs" by I. Moerdijk and J. van Oosten to have some basic understanding of set theory, when I hit this paragraph:

... Bases for vector spaces. Let $V$ be a vector space over
  $\mathbb{R}$ (or, in fact, any other ﬁeld), for example the set of
  continuous functions from $[0, 1]$ into $\mathbb{R}$. Then $V$ has a
  basis, that is a subset $B \subset V$ with the property that every 
  $v\in V$ can be written as a ﬁnite sum $$v = k_1b_1 + \dots + k_nb_n$$
  with $k_1,... ,k_n \in \mathbb{R}$ and $b_1,... ,b_n \in B$, and
  moreover this ﬁnite sum is unique.

I'm not sure here -- how to guarantee the sum is finite?
Take the example that vector space $V$ is the set of continuous functions from $[0, 1]$ into $\mathbb{R}$. then as I remember $\sin(n\pi x)$ and $\cos(n\pi x)$ together is a set of basis -- they are used in Fourier transform.
However, for a particular function $v\in L^1[0,1]$, say $f(x)=x^2$, it's not possible to decompose it in finite sum of $\sin(n\pi x)$ and $\cos(n\pi x)$.
And we don't care about this as after enough items, the approximation is quite close.
So why the paper mentions the sum is finite?
 A: There are two major notions for a basis for a vector space.
There is a Hamel basis, or a linear basis, which means that every vector is the sum of finitely many basis elements; and there is a Schauder basis, or a topological basis, which means that the span of the basis is dense in the vector space.
Schauder bases make sense in the case where we have a topology (e.g. a normed vector space), and Hamel bases always make sense.
We can show, for example, that $\ell^2$ has a Schauder basis quite easily, and your argument is exactly why Fourier bases are topological bases, and not Hamel bases. But the axiom of choice guarantees us that there is a Hamel basis as well.
It should be noted that the Axiom is equivalent to the statement that every vector space has a Hamel basis, so in most cases this basis is intangible. In particular, in the case of function spaces we can show some models of set theory in which the axiom of choice fails and they have no Hamel bases (but of course they still have Schauder bases).
