It is the set of all the function $\mathbb{R}\to \mathbb{R}$.
This is true in general for every $X^Y$ (as noted in the comments by Lord Shark the Unknown), and somewhat follows from $X^Y:=\Pi_{Y}X$ (the idea here is that one of the possible equivalent definition of a cartesian product $\Pi_{\mathcal{F}}X_f$ is the family of function from that associates $f\in \mathcal{F}$ with some element of $X_{f}$. The existence of such a function, called choice function, is highly non trivial and follows from the axiom of choice as you can see here).
Now, I claim that $\mathbb{R}^\mathbb{R}$ is not finitely generated. To prove it, let $\mathbb{P}$ be the vector subspace of all the polynomials. Now, given any finite l.i. set of vectors in $\mathbb{P}$, their products cannot be obtained as a linear combination of those vectors. Thus, $\mathbb{P}$ is not finitely generated, and so $\mathbb{R}^\mathbb{R}\supset \mathbb{P}$ cannot be finitely generated either.
Thus, $A$ is not a basis of $\mathbb{R}^\mathbb{R}$, as neither is any finite set of vectors.
Note: $\mathbb{R}^\infty$ is usually used to denote $\mathbb{R}^\mathbb{N}$, and so is not an equivalent notation