Is this a valid proof that $0.\overline{9} = 1$? There
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This is something that intrigues me, and I have a question about this, including a "proof" which I have found myself.
Question:
This comment says that

you need a terminating decimal to get something less than 1.

If so, does it mean that a non-terminating decimal (e.g. $0.\overline{9}$) is $\ge 1$?
So is $\frac{1}{3}$ ($0.\overline{3}$) also $\ge 1$? It is non-terminating, but you can subtract $\frac{1}{3}$ from $1$ to get $\frac{2}{3} = 0.\overline{6}$, which is another non-terminating decimal. How do those mechanics work?

Theorem: $0.99999... = 1(.00000... = 1)$
Proof:
\begin{align}
\frac{1}{9} &= 0.11111... \\
\frac{2}{9} &= 0.22222... \\
\frac{3}{9} &= 0.33333... \\
\frac{4}{9} &= 0.44444... \\
\frac{5}{9} &= 0.55555... \\
\frac{6}{9} &= 0.66666... \\
\frac{7}{9} &= 0.77777... \\
\frac{8}{9} &= 0.88888... \\
\therefore \frac{9}{9} &= 0.99999... \\
&= 1
\end{align}
Is the above proof correct? I came up with it myself before I decided to ask this question, but I do not know if it is mathematically valid.

The non-terminating notation (either $0.9999\cdots$ or $0.\overline9$) is a disguised limit, namely
$$\lim_{n\to\infty}\sum_{k=1}^n\frac 9{10^k}$$ or $$\lim_{n\to\infty}\left(1-10^{-n}\right).$$
This limit equals $1$.

To relate this to your proof, we indeed have
$$0.\overline1=\lim_{n\to\infty}\sum_{k=1}^n\frac 1{10^k}=\frac19,$$
then
$$9\cdot 0.\overline1=9\lim_{n\to\infty}\sum_{k=1}^n\frac 1{10^k}=\lim_{n\to\infty}\sum_{k=1}^n\frac 9{10^k}=0.\overline 9$$
and $$\frac99=1.$$
But honestly, I see no benefit taking this indirect route through $9\cdot0.\overline 1$, and for completeness, you should explain why $\dfrac19=0.\overline1$, and why $9\cdot0.\overline1=0.\overline9$ like I did (or another way).
