In Beals' Analysis: an Introduction, there is a proof that L2 is complete. I follow the proof, except for one last comment. I will sketch the proof here for context:
Given a Cauchy sequence $(f_n)_{n=1}^\infty$, choose a subsequence $(g_n)_{n=1}^\infty$ such that
$$\|g_{n+1} - g_n \|_2 \leq 2^{-n}. $$
For a bounded interval $A$, we have by Cauchy Schwartz,
$$\int_A |g_{n+1}-g_n| \leq \|g_{n+1} - g_n\| m(A)^{1/2}.$$
It follows that $f(x) = \lim g_n(x)$ exists almost everywhere in $A$, so it also exists almost everywhere in $\mathbb R$.
Now, we need to show that $\|g_n -f\|_2 \rightarrow 0$, and let $h$ be any element of $L^2$. Then,
$$[f(x) - g_n(x)] h(x) = \sum_{k=n}^\infty [ g_{k+1}(x) - g_k(x) ]h(x)$$
$$|(f-g_n,h) | \leq \sum_{k=n}^\infty \|g_{k+1} - g_k \|_2 \|h\|_2 \leq 2^{1-n} \|h\|_2$$
We let $h=f-g_n$, and then
$$\|f-g_n\|_2^2 \leq 2^{1-n} \| f-g_n\|_2$$
so $\|f-g_n\|_2 \leq 2^{1-n}$.
Which completes the proof.
The author remarks, though, that "Strictly speaking, we should let $h$ run through an appropriate $f-g_n$, since we do not know ahead of time that $f-g_n$ is actually square integrable."
Indeed, I recognize that to be a serious problem. It might not even make sense to write $\|f-g_n\|_2$. I'm just not sure how running through an appropriate sequence solve the problem. Can you help me make this explicit?
Here is one standard argument:
Choose $N$ such that $\|g_n-g_m\| \le 1$ for all $n,m \ge N$.
Note that $\liminf_n g_n(x)-g_N(x)| = |f(x)-g_N(x)|$ for ae. $x$.
Fatou gives $\int |f(x)-g_N(x)| \le \liminf_n \|g_n-g_N\| \le 1$.
Hence $f-g_N$ is in $L^2$ and hence so is $f = (f-g_N)+g_N$.