I don't need induction because I am not proving a novel statement for all n, but rather for one specific n. You already agree with me that the partial sums are of the form $$1 - \frac{1}{10^k}$$ so I don't need to prove that anew with induction.

You aren't even using the correct domain of mathematics. This thread is about the real numbers and analysis.

$$y = 10^x$$ is continuous and one-to-one mapping all real numbers to positive real numbers, therefore $$x = \log_{\tiny 10} y$$ is continuous and maps all positive numbers to the real numbers. This is very basic analysis.

Therefore if $$0 \lt 0.999... \lt 1$$ then $$1 - 0.999...$$ is positive and therefore $$\log_{\tiny 10} \left( 1 - 0.999... \right)$$ is a real number less than 0, therefore $$ \left\lfloor \log_{\tiny 10} \left( 1 - 0.999... \right) \right\rfloor $$ is an negative integer and $$1 - \left\lfloor \log_{\tiny 10} \left( 1 - 0.999... \right) \right\rfloor$$ is a positive integer, and therefore a natural number. We call this natural number n.

Now because we assumed $$ 0.999... \lt 1$$, we know n exists and that $$0.999... \lt 1 - \frac{1}{10^n}$$ but because $$1 - \frac{1}{10^n} \in \left{ 0, \, \frac{1}{10}, \, \frac{1}{10^2}, \, \frac{1}{10^3}, \, \dots \right}$$ then and $$0.999... \; = \; \textrm{sup} \, \left{ 0, \, \frac{1}{10}, \, \frac{1}{10^2}, \, \frac{1}{10^3}, \, \dots \right}$$ it follows that necessarily, $$1 - \frac{1}{10^n} \leq 0.999...$$ and from the transitive axiom of ordering, it follows that $$0.999... \lt 0.999...$$ which contradicts $$0.999... = 0.999...$$ and so we reject the hypothesis $$ 0.999... \lt 1$$.

At no point did I mean to write $$\color{red} \textrm{sup} \, \left{ 0, \, \frac{1}{10}, \, \frac{1}{10^2}, \, \frac{1}{10^3}, \, \dots \right} \lt 0.999... $$ because that contradicts what I am asserting $$0.999... = \textrm{sup} \, \left{ 0, \, \frac{1}{10}, \, \frac{1}{10^2}, \, \frac{1}{10^3}, \, \dots \right}$$ which in layman's terms is 0.999... is the smallest number at least as big as any number of the form $$1 - \frac{1}{10^k}$$ where k is any natural number. I see I did do it as a typo,

Obviously, $$1 - \frac{1}{10^{n-3}} \lt 0.999... \lt 1 - \frac{1}{10^n} \lt \textrm{sup} \; \left{ 0, \, 1 - \frac{1}{10}, \, 1 - \frac{1}{10^2}, \, 1 - \frac{1}{10^3}, \dots \right} = 0.999...$$ was meant. But it doesn't matter much since both $$1 - \frac{1}{10^n} \lt \textrm{sup} \; \left{ 0, \, 1 - \frac{1}{10}, \, 1 - \frac{1}{10^2}, \, 1 - \frac{1}{10^3}, \dots \right} $$ and $$1 - \frac{1}{10^n} \lt 0.999...$$ are true for any natural number n, and since by hypothesis $$0.999... \lt 1 - \frac{1}{10^n}$$ then $$0.999... \lt 0.999... \lt 0.999... $$ and it follows that $$0.999... \lt \textrm{sup} \; \left{ 0, \, 1 - \frac{1}{10}, \, 1 - \frac{1}{10^2}, \, 1 - \frac{1}{10^3}, \dots \right} \lt 0.999... $$. That one mistaken hypothesis (assuming $$ 0.999... \lt 1$$) leads to more than one faulty conclusion is called the Principle of Explosion.

Whether or not I was consciously meaning to assert that explicitly, I do not recall. But I didn't call it out in text, and that is a mistake of sorts.