Show that the vector space of real functions is not finitely spanned Here's what I've come up with. Let $\mathcal B$ be a generating set of $\mathcal F(R)$. Let us assume that $\mathcal B$ is a finite set, thus: $\lt  \mathcal B \gt = \lt \{b_1, ..., b_n \}\gt, n \in N$. Let  $\mathcal f \in F(R).$ Thus: $\mathcal f = \alpha_1 b_1 + ... + \alpha_nb_n,$ where $ \alpha_1,..., \alpha_n \in R$ are not all zeroes.
If  $\mathcal F(R)$ is a vector space, so there can be a function  $\mathcal g \in F(R)$ such as  $\mathcal g \neq \lambda f, \lambda \in R$. Thus  $\mathcal f+g \in F(R)$  but  $\mathcal f+g \neq \alpha_1 b_1 + ... + \alpha_nb_n$. If is $\mathcal g$ is to belong to $ \lt \mathcal B \gt$ we must add a new vector to $\mathcal B$.
Repeating the above mentioned procedure, we can find a function $\mathcal g\in F(R)$ such as $\mathcal h \neq \mathcal\lambda g$. Thus, we can find infinite functions and thus need infinite base vectors.
Is this proof right?

I think I understand the spirit of your proof, but it doesn't seem rigorous.
What you are trying to prove is equivalent to saying that $\mathcal{F}(R)$ is an infinite-dimensional vector space (if $\mathcal{F}(R)$ were finite dimensional, any basis would be a finite spanning set). Therefore if we could display an infinite collection of linearly independent vectors, we'd be done.
Can you show that the set $\{1, x, x^2, \dots \}$ is linearly independent?
