Every Hilbert space is isometrically isomorphic with $\ell^2$ 
Let $H$ be a hilbert space and let $\{u_\alpha\}_{\alpha \in A}$ be a
  orthornormal basis ($A$ is not supposed to be countable a priori).
  Then there is an isometric isomorphism between $H$ and
  $\ell^2(\mathcal{A})$ given by $x \mapsto \hat{x}$ with $\hat{x}_\alpha = \langle x,u_\alpha\rangle$.

Proof:
Linearity follows by linearity of scalar product: $(\widehat{ax + by})_\alpha = \langle ax + by,u_\alpha\rangle = a\langle x,u_\alpha\rangle + b\langle y,u_\alpha\rangle = a\hat{x}_\alpha + b\hat{y}_\alpha$
From Parseval we have $\|x\| = \sum_{\alpha \in A} |\langle x, u_\alpha\rangle| = \|\hat{x}\|_2$ and then the mapping is an isometry (and hence also injective).
To conclude we need to prove that this isometry is also surjective.
Let then $f$ be in $\ell^2(A)$. Then, by definition of $\ell^2$, $\sum_{\alpha \in A} |f_\alpha|^2 < \infty$, this implies that $|\{\alpha : f_\alpha \not = 0 \}| \le |\mathbb{N}|$ and thus also the sum $\sum_{\alpha \in A} f_\alpha u_\alpha$ has countable many nonzero terms. Then, by Pythagorean theorem (using continuity of inner product) $\|\sum f_\alpha u_\alpha \|^2 = \sum_{\alpha \in A} |f_\alpha|^2 < \infty$ so this series is absolutely convergent and then $z = \sum_{\alpha \in A} f_\alpha u_\alpha$ exists. Clearly $\hat{z} = f$ 
$\square$
Q1: Is this proof correct?

This proof is worked out from Folland's Real Analysis book. In the book though, Prof. Folland writes:

If $f \in \ell^2(A)$ then $\sum_{\alpha \in A} |f_\alpha|^2 < \infty$
  so the Pythagorean theorem shows that partial sums of the series
  $\sum_{\alpha \in A} f_\alpha u_\alpha$ (of which only countable many
  terms are nonzero) are Cauchy.

I know the definitions but I don't get why Prof. Folland says that Pythagorean theorem asserts that partial sums are Cauchy. I think Pythagorean theorem in this case give us absolute convergence of the sum, not just the fact partial sums are Cauchy. In this case the space is complete so there is no real distinction between Cauchy and convergence but I care about logical order of the proof.
Q2: Why Pythagorean theorem asserts only Cauchy property of partial sums and not absolute convergence as I said?
Thanks in advance

Note: Uncountable sums of non negative terms are defined like this:
$$ \sum_{\alpha \in A} g(\alpha) := \sup \left\{ \sum_{\alpha \in F} g(\alpha) : F \subset A, \ F \ \text{finite} \right\}$$
with $g: A \to [0,\infty)$

 A: If $\mathcal{H}$ is a Hilbert space and $\{ e_{\alpha} \}_{\alpha\in\Lambda}$ is an orthonormal subset on $\mathcal{H}$, then
$$
     x=\left(x-\sum_{n=1}^{N}(x,e_{\alpha_n})e_{\alpha_n}\right)+\sum_{n=1}^{N}(x,e_{\alpha_n})e_{\alpha_n} \;\;\; (\dagger)
$$
is an orthogonal decomposition. Therefore,
\begin{align}
          \|x\|^2&=\|x-\sum_{n=1}^{N}(x,e_{\alpha_n})e_{\alpha_n}\|^2+\|\sum_{n=1}^{N}(x,e_{\alpha_n})e_{\alpha_n}\|^2 \\
        &\ge \|\sum_{n=1}^{N}(x,e_{\alpha_n})e_{\alpha_n}\|^2 \\
        &= \sum_{n=1}^{N}|(x,e_{\alpha_n})|^2
\end{align}
Therefore, for any $\epsilon > 0$, the following set is finite:
$$
             \Lambda_{x,\epsilon}=\{ \alpha \in \Lambda : |(x,e_{\alpha})| \ge \epsilon \}
$$
Therefore, the following set is empty, finite, or countable infinite.
$$
           \{ \alpha\in \Lambda : (x,e_{\alpha}) \ne 0 \} = \bigcup_{N=1}^{\infty}\Lambda_{x,1/N}.
$$
Let $x\in\mathcal{H}$, and let $\{\alpha_n\}$ be an enumeration of the $\alpha\in\Lambda$ for which $(x,e_{\alpha})\ne 0$. The only interesting case is where this set of indices is countably infinite. In that case, the original orthogonal decomposition $(\dagger)$ gives
$$
         \|x\|^2=\|x-\sum_{n=1}^{N}(x,e_{\alpha_n})e_{\alpha_n}\|^2+\sum_{n=1}^{N}|(x,e_{\alpha_n})|^2.
$$
Therefore $\|x\|^2=\lim_{N}\sum_{n=1}^{N}|(x,e_{\alpha_n})|^2$ iff
$$
            \lim_{N}\|x-\sum_{n=1}^{N}(x,e_{\alpha_n})e_{\alpha_n}\|=0.
$$
So, Parseval's equality holds for some $x$ with respect to $\{ e_{\alpha} \}_{\alpha\in\Lambda}$ iff $\lim_{N}\sum_{n=1}^{N}(x,e_{\alpha_n})e_{\alpha_n}=x$ converges in the norm of $\mathcal{H}$. And, in that case, any ordering for the sum will give the same limit because the Parseval equality involves an absolutely convergent series of positive numbers and, hence, may be ordered in any way. This is what is meant by the unordered sum converging.
