Background: I'm studying the properties of summability within orthogonal (possibly uncountable) families of vectors and have come across the following theorem (Theorem 8.2 of "Introduction to Hilbert Space and the Theory of Spectral Multiplicity" by P. R. Halmos).
Theorem: An orthogonal family of vectors $\left\{ x_{j} \right\}$ is summable if and only if the family $\left\{ \| x_{j} \|^{2} \right \}$ of positive numbers is summable; this condition may also be expressed by writing $\Sigma_j\left\|x_j\right\|^2<\infty$. If $x=\Sigma_j x_j$, then $\|x\|^2=\Sigma_j\left\|x_j\right\|^2$.
Proof: $\ldots$
The second assertion of the theorem is a consequence of the relations $$ \|x\|^{2} = \langle x, x\rangle = \left\langle \Sigma_j x_{j}, x \right \rangle = \Sigma_{j} \left \langle x_{j}, x \right \rangle = \Sigma_{j} \left\langle x_{j}, \Sigma_k x_{k}\right \rangle = \Sigma_j \Sigma_k \left\langle x_{j}, x_{k}\right\rangle = \Sigma_j \left\langle x_{j}, x_{j} \right\rangle = \Sigma_j \| x_{j} \|^{2} . $$ (The second to last step in this chain of equations depends on the obvious fact that if all but one of the terms of a family of vectors, or, in particular, of complex numbers, are zero, then that family is summable and its sum is the exceptional term.)
Question: Am I right that the author implicitly makes use of the fact that the convergence of an uncountable sum implies that at most countable many elements are non-zero? Hence the above sums are all over countable index sets?
To be more explicit if $J=\mathbb{N}$ the chain of equations follows by exploiting linearity and continuity of the inner product. F.e. in that case, the second "$ = $" involves the following steps which are usually hidden: $$ \left \langle x, x \right \rangle = \left \langle \sum_{j=1}^{\infty}x_{j}, x \right \rangle = \left \langle \lim_{n \to \infty} \sum_{j=1}^{n}x_{j}, x\right \rangle = \lim_{n \to \infty} \left \langle \sum_{j=1}^{n}x_{j}, x \right \rangle = \lim_{n \to \infty} \sum_{j=1}^{n} \left \langle x_{j}, x \right \rangle = \sum_{j=1}^{\infty} \left \langle x_{j}, x \right \rangle $$
Definition: A family $\{x_{j}\}$ of vectors will be called summable with sum $x$ in symbols $\sum_{j}x_{j} = x$, if for every positive number $\epsilon$ there exists a finite set $J_{0}$ of indices such that $\| x - \sum_{j \in J}x_{j} \| < \epsilon$ whenever $J$ is a finite set of indices containing $J_{0}$.
The fact that the convergence of an uncountable sum implies that at most countable many elements are non-zero is useless here. You can do the same steps with summable families, as follows.
Let us assume $\sum_jx_j=x$ and prove that $\|x\|^2=\sum_j\left\|x_j\right\|^2.$
By hypothesis, for every positive number $\epsilon$, there exists a finite set $J_0$ of indices such that $\| x - \sum_{j \in J}x_j\| < \epsilon$ whenever $J$ is a finite set of indices containing $J_0$.
For $\epsilon,J$ as above, and $y:=\sum_{j \in J}x_j$, we have $$\begin{align}\|x\|^2&=\langle x,x-y\rangle+\langle x,y\rangle\\ &=\langle x,x-y\rangle+\langle x-y,y\rangle+\|y\|^2.\end{align}$$ By Cauchy-Schwarz and triangular inequalities, $$\left|\langle x,x-y\rangle+\langle x-y,y\rangle\right|\le\epsilon(\|x\|+\|y\|)\le\epsilon(2\|x\|+\epsilon).$$ Moreover, since the $x_j$'s are pairwise orthogonal, $\|y\|^2=\sum_{j\in J}\|x_j\|^2.$
We thus proved that $\forall\epsilon>0,\exists J_0$ finite, $\forall J$ finite $\supset J_0,$ $$\left|\|x\|^2-\sum_{j\in J}\|x_j\|^2\right|\le\epsilon(2\|x\|+\epsilon).$$
To conclude, it suffices to notice that $\forall\eta>0,\exists\epsilon>0,\epsilon(2\|x\|+\epsilon)<\eta$, i.e. $\epsilon\in\left(0,-\|x\|+\sqrt{\|x\|^2+\eta}\right).$