Let $V$ be an infinite dimensional Hilbert space and $W$ its dual. Can I identify a $w \in W$ by how it acts on a basis, $B$, of $V$? Does it matter if the dimension of $V$ is uncountable and/or if it's a Banach space? Does it matter if we make $W$ the continuous dual?
I know this works for finite dimensions but I'm not sure for infinite dimensions since they aren't spanned by the basis.
The word "basis" is ambiguous. In the case of Hilbert spaces, most likely you are talking about the orthonormal basis. In the case of Banach spaces, there is no general notion of "basis", except the Hamel one. But in all cases I know of, a basis is always a subset $B$ of the space $X$, such that the topological closure of the linear span of $B$ is $X$.
Now, given a "basis" $B$ of $X$, and an element $v$ from $X^*$ -- the continuous dual, by linearity the action of $v$ on the linear span of $B$ is determined by its action on $B$, and then by continuity, the action of $v$ on $X$ is completely determined as well, since $\operatorname{Span}(B)$ is dense in $X$. In particular, even if a Hilbert space is non-separable, the argument still works.
However, unless $B$ is the Hamel basis, we do need the continuity to conclude. Indeed, if $B$ is not the Hamel basis, then it's easy to show (by Zorn's lemma) that we may find subspace $Y$ such that $X=\operatorname{Span}(B)\oplus Y$ (a pure linear algebra decomposition), and two functionals agreeing with each other on $B$ only guarantees they agree on $\operatorname{Span}(B)$, not necessarily over $Y$, hence not $X$ either. To be more rigorous, we may pick $y\in X\setminus\operatorname{Span}(B)$, and using Zorn's lemma, we may construct two linear functionals $v_1, v_2$ such that $v_1|_B=v_2|_B=0_B$ and $v_1(y)=0, v_2(y)=1$.
