 Solution

Use the convention that the inner product is linear in its first argument. Set $a_n=⟨x,e_n⟩$ and $P_{N}x={\sum }_{n\le N}{a}_n{e}_n$. Orthogonality makes $P_{N}x$ the nearest point in the finite span of the first $N$ basis vectors. Since an orthonormal basis has dense linear span, for every $ϵ>0$ a finite linear combination approximates $x$ within $ϵ$. For all larger $N$, the nearest-point property gives $\Vert x-P_{N}x\Vert <ϵ$. Thus the series converges in $X$ to $x$ even without completeness. Taking inner products with each $e_n$ proves uniqueness of its coefficients.

In the Hilbert-space case, Parseval's identity gives $\sum |a_n{|}^2=\Vert x{\Vert }^2$. Completeness makesconvergent. It is linear and preserves inner products and norms, so bounded with norm one. Its inverse is the corresponding map sending $f_n$ to $e_n$; equivalently every coefficient sequence of a vector in the $f_n$ basis is also square-summable in the $e_n$ basis. Hence $U$ is surjective and unitary. Any bounded map agreeing on the $e_n$ agrees on their dense span and then on all of $X$, proving uniqueness.