Riesz-Fischer theorem The aim of this exercise is to prove the Riesz-Fischer theorem for Hilbert spaces that aren't separable.

Let $I$ an index set and $1\leq p \leq \infty$. Let
   $\mathcal{F}=\{F\subset I: F$  is finite$\}$. Given $(a_i)_{i\in
>  I}\subset \mathbb{K}$, we define:
$$\Vert (a_i)_{i\in I} \Vert_p = \sup_{F\in\mathcal{F}}\left( \sum_{i\in F}
|a_i|^p\right)^{1/p} \qquad \mbox{and} \qquad \Vert (a_i)_{i\in
I}\Vert_\infty = \sup_{i\in I}|a_i| $$
  
  
*
  
*Prove that $l_p(I)=\{ (a_i)_{i\in I}\} \subset \mathbb{K}:\Vert (a_i)_{i\in I} \Vert_p <\infty  \}$ is a Banach space with the norm $\Vert\cdot\Vert_p$.
  
*$l_p(I)'$ is isometrically isomorphic to $l_q(I)$ where $1/p+1/q = 1$.
  
*Given a orthonormal basis $S=\{x_i: i\in I\}$ for a Hilbert space $H$, prove that H is isometrically isomorphic to $l_2(I)$.
  
*Prove that every Hilbert Space $H$ is isometrically isomorphic to $H'$.

The second item is quite the same proof of $ l_p (\mathbb{N})$, and the 4 follows immediately from 3. I'm having problems to prove item 3.
The map $T:H \longrightarrow l_2(I)$ defined by $T(x)=( \langle x, x_i\rangle)_{i\in I}$ is a linear isometry. I need help to prove that T is surjective, then I can use the open map theorem and conclude the proof. :
Let $y=(a_i)_{i\in I} \in l_2(I)$, consider $\sum_{i\in I} a_ix_i$. How can I show that this sum converge to an element of $H$? (I used summability, but I'm not satisfied with my argument, there is a proof that don't requires summability?)
 A: The first question is, in what sense do you want to understand convergence of the sum?
The correct notion here is unconditional convergence, i.e. we show that there is some $h \in H$ such that for every $\epsilon > 0$, there is some finite subset $J_\epsilon \subset I$ with $\Vert h - \sum_{j\in J} a_j x_j \Vert < \epsilon$ for all finite sets $J_\epsilon \subset J \subset I$.
To this end, first note $a_i = 0$ for all but countably many $i$ (because of $(a_i)_i \in \ell^2$). Hence, let $(i_n)_n$ be pairwise distinct with $\{i_n \mid n\} = \{i \mid a_i \neq 0\}$ (if there are only finitely many, the claim is trivial).
Consider the sequence $h_N := \sum_{n=1}^N a_{i_n} x_{i_n}$. We then have (for $N \geq M \geq N_0$)
$$
\Vert h_N - h_M\Vert^2 = \Vert \sum_{n=M+1}^N a_{i_n} x_{i_n}\Vert ^2 = \sum_{n=N+1}^M |a_{i_n}|^2 \leq \sum_{n=N_0 + 1}^\infty |a_{i_n}|^2 \xrightarrow[N_0 \to \infty]{} 0
$$
where we used orthogonality of the $x_i$ and Pythagoras theorem in the step before the last one.
Hence, the sequence $(h_N)_N$ is Cauchy and thus convergent to some $h \in H$ by completeness of $H$.
It remains to prove that we indeed have unconditional convergence to this "candidate limit". To see this, let $\epsilon >0$ be arbitrary. Then there is a finite $J\epsilon \subset I$ with $\sum_{i \in I\setminus J_\epsilon}|a_i|^2<\epsilon$. Now let $J_\epsilon \subset J \subset I$ be finite. We get
$$
\Vert h-\sum_{j \in J} a_j x_j\Vert ^2=\lim_N \Vert \sum_{n=1}^N a_{i_n}x_{i_n}-\sum_{j\in J} a_j x_j\Vert^2=\lim_N \sum_{j \in J \Delta \{i_n \mid n=1\dots N} |a_j|^2,
$$
where we again used Pythagoras theorem. Here, $\Delta$ denotes the symmetric difference. 
Because of $\{i_n \mid n\in \Bbb{N}\}=\{i\mid a_i \neq 0\}$, the above limit is equal to
$$
\sum_{i \in I \setminus J}|a_i|^2<\epsilon,
$$
which completes the proof. 
