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 claimed continuity is responsible for proofs regarding natural numbers. Continuity requires the real number system. You cannot derive continuity from only natural numbers. So, you are wrong here.

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,

This follows from continuity and monotonicity of the exponential because $$\forall k\in \mathbb{Z} \; \frac{1}{10^{k+1}} \lt \frac{1}{10^k} $$ and so if $$0 \lt 0.999... \lt 1$$ then $$n \quad = \quad 1 - \left\lfloor \log_{\tiny 10} \left( 1 - 0.999... \right) \right\rfloor \quad \in \mathbb{N}$$ and therefore $$1 - \frac{1}{10^{n-3}} \lt 0.999... \lt 1 - \frac{1}{10^n} \lt \textrm{sup} \; \left{ 0, \, 1 - \frac{1}{10}, \, 1 - \frac{1}{10^2}, \, 1 - \frac{1}{10^3}, \dots \right} $$$$\color{red} \lt$$$$ 0.999...$$ which is a contradiction that proves $$0.999... \not\lt 1$$.

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.