I keep confusing myself about a subspace basis and I can only find intelligible material discussing the finite, linear algebra, case.
It is known that the Hilbert space $L^2(X)$ has a basis, for example given by Fourier modes when $X=[-L,L]$. Consider the vector subspace $S=L^2(X)\cap L^4(X) \subset L^2(X)$.
What is the basis of $S$?
A weaker, but still relevant to me, question is does $S$ have a basis without assuming the axiom of choice?
I think that you are confusing basis with complete orthonormal set.
The Fourier modes $\{e^{inx}\}_{n\in\mathbb{Z}}$ are a complete orthonormal set in $L^2[-\pi,\pi]$. This means that for any $f\in L^2$, its Fourier series converges in the $L^2$-norm to $f$. But it is not a basis of $L^2[-\pi,\pi]$. $L^4$ is not a Hilbert space, and it does not make sense to ask for an orthonormal set of $L^4$.
A basis is an algebraic concept. A basis of $L^2[-\pi,\pi]$ is a set of functions $\{f_j\}_{j\in J}$ indexed by a set $J$ such that for any $f\in L^2$, there is a finite set of indices $j_1,\dots,j_k\in J$and scalars $c_1,\dots,c_k$ such that $f=\sum_{i=1}^k c_i\,f_{j_i}$. A basis for $L^2\cap L^4$ will be $\{f_j:j\in J\text{ and } f_j\in L^4\}$.
The Fourier exponentials $\{e^{inx}\}_{n\in\mathbb{Z}}$ are a Schauder basis (not a Hamel basis) of $L_2([-\pi,\pi])$. In fact they are a Schauder basis for $L_p([-\pi,\pi])$ for any $1<p<\infty$.
Unlike Hamel bases, Schauder bases allow for countable combinations.
The term basis is commonly used to refer to a complete orthonormal set in $L_2$, since it is a Schauder basis.
– Dunham Oct 28 '14 at 13:52Wojtaszczyk, Banach spaces for analysts (see p.40)
Young, An introduction to non-harmonic Fourier series