Prove that $l^2(I)$ is a Hilbert Space. This problem is taken from Conway's A Course in Functional Analysis section 1.1 problem 2
Verify that $l^2(I)$ (Example 1.7) is a Hilbert Space.
Below is my attempt at a solution. Any feedback is greatly appreciated!
To show completeness, suppose we had a Cauchy sequence $\{x_n\}\in l^2(I)$. We see that:
\begin{align}
    ||x_n - x_m||^2 =\sum_{i} ||(x_n(i) - x_m(i)||^2    
    \end{align}
So that $||x_n - x_m|| \rightarrow 0$ as $m,n \rightarrow \infty$ must also imply $||x_n(i) - x_m(i)|| \rightarrow 0$
Thus, $x_n(i)$ is a Cauchy sequence over the real/complex numbers, we it must converge to a limit,  say $x(i)$. Let us construct $x(i)$ this way for all $i \in I$ and show that $x(i) \in l^2(I)$.
We first show that $x(i) = 0$ for all but a countable number of $i$. Suppose the contrary. Then for any $x_n$, $||x - x_n||^2$ being a sum of uncountably many nonzero elements, which cannot be bounded. The contradiciton proves the desired result.
Now We try to show that $\sum_{i\in I}|x(i)|^2 < \infty$. But we know that for any $\epsilon>0$, there is sufficiently large $n$ such that we have $||x - x_n|| < \epsilon$. Thus
\begin{align}
        \sum_{i \in I}|x(i)|^2 &= ||x||^2\\
        &= ||x - x_n + x_n||^2
        \leq \left(||x-x_n|| + ||x_n||\right)^2 \quad \text{ by triangle inequality}\\
        &< \left(\epsilon + ||x_n||\right)^2\\
        &< \infty
    \end{align}
where we noted in the last line that $||x_n||< \infty$ since $x_n \in l^2(I)$. Thus we have shown $x \in l^2(I)$ as desired.

You don't know that $\lVert x - x_n\rVert < +\infty$ until you have shown that $x \in \ell^2(I)$.
The trick is to take a finite $F \subset I$, then $$\sum_{i \in F} \lvert x(i)\rvert^2 = \lim_{n \to \infty} \sum_{i \in F} \lvert x_n(i)\rvert^2 \leqslant M := \sup_n\: \lVert x_n\rVert < +\infty\,.$$
Since the bound $M$ is independent of $F$, it follows that
$$\sum_{i \in I} \lvert x(i)\rvert^2 \leqslant M\,,$$
hence $x \in \ell^2(I)$. Then it remains to show that $\lVert x - x_n\rVert \to 0$, which we do with the same trick,
$$\sum_{i \in F} \lvert x(i) - x_n(i)\rvert^2 = \lim_{m \to \infty} \sum_{i \in F} \lvert x_m(i) - x_n(i)\rvert^2 \leqslant \limsup_{m \to \infty} \lVert x_m - x_n\rVert^2\,.$$
This again holds for all finite $F \subset I$ with the bound not depending on $F$, hence
$$\lVert x - x_n\rVert^2 \leqslant \limsup_{m \to \infty} \lVert x_m - x_n\rVert^2\,.$$
Since $(x_n)$ is a Cauchy sequence, the right hand side tends to $0$ as $n \to \infty$.
