This is a follow-up of this post.
Here is the statement we want to prove.
Let $\{x_j\}$ be an arbitrarily indexed orthonormal set in a Hilbert space(possibly non-separable). Show that the closed linear span of $\{x_j\}$ consists of all vectors of the form:$$x=\sum a_j x_j$$ where the sum converges in the sense of Hilbert space norm and the $a_j$ are complex number so chosen that$$\sum \lvert a_j \rvert^2 < \infty$$ Furthermore:$$\lVert x\rVert^2=\sum \lvert a_j \rvert^2$$ with $a_j=(x,x_j)$. (circular brackets denotes the inner product)
My question is the following. Suppose that $y\in\overline{\text{span}\{x_j\}}$. Then there exist a sequence $y^{(n)}\in\text{span}\{x_j\}$ such that $y^{(n)} \to y$. Then there exists a countable index set $I$ such that \begin{align} y^{(n)} = \sum_{i\in I} a^{(n)}_{i}x_{j_i} \end{align} for each $n$. How can I prove that there exists scalars $a_{i}$ such that \begin{align} y = \sum_{i\in I} a_{i}x_{j_i} \end{align} and \begin{align} \sum_{i\in I} |a_{i}|^2 < \infty, \end{align} as claimed in an answer?