I'm wondering is this proof will fix the issue of being "completely fallacious" - or is it still circular?
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Consider the function $f(x)=x^2.$ The aim is to prove its continuity using interval notation. Let $a\in \mathbb{R}.$ We need to check that given $\epsilon>0$ there exists $\delta>0$ such that for all $x$ satisfying $a-\delta<x<a+\delta$ one has $a^2-\epsilon<x^2<a^2+\epsilon.$ To understand what must be the value of $\delta,$ we observe that $$ x^2-a^2=(x-a)(x+a). $$ If we know that $|x-a|<\delta,$ then $|x|<\delta+|a|$ and $$ |x^2-a^2|=|x-a||x+a|<\delta(|x|+|a|)<\delta(2|a|+\delta). $$ Take $\delta=\min(1,\frac{\epsilon}{2a+1})$ and let $a-\delta<x<a+\delta.$ Then $$ x^2=a^2+x^2-a^2<a^2+\delta(2|a|+\delta)\leq a^2+\delta(2|a|+1)\leq a^2+\epsilon $$ and $$ x^2=a^2+x^2-a^2>a^2-\delta(2|a|+\delta)\geq a^2-\delta(2|a|+1)\geq a^2-\epsilon $$ Finally, if $a-\delta<x<a+\delta,$ then $a^2-\epsilon<x^2<a^2+\epsilon.$
The proof that you wrote down, completely correctly, is the standard one, and it uses the "factoring" mentioned in the cited text that the "fallacious argument" was thought to avoid. The "factoring" is exactly the use of the binomial formula $$ x^2-a^2=(x-a)(x+a)\\ x^3-a^3=(x-a)(x^2+xa+a^2) $$ that stands at the start of your proof.
Just any property of the square root function $\sqrt{x}=\sup\{r\in\Bbb R_+:r^2\le x\}$, that it is well-defined, monotonous, continuous, makes use of the same properties of the square function which are proven starting from this factorization.