Seeking feedback on my $\epsilon$-$\delta$ proof of $\lim_{x \to a} x^2 = a^2$.

I'm seeking feedback on my understanding of the $\epsilon$-$\delta$ limit proof for quadratic functions, specifically for $\lim_{x \to a} x^2 = a^2$.After studying multiple proofs, I've noticed that while the general approach is similar, the justifications for certain steps often vary. Some explanations, such as "because $\delta$ is typically small, we suppose $\delta \leq 1$", don't seem entirely rigorous or satisfying.

Nevertheless, these various proofs have helped me grasp the overall idea. I've attempted to synthesize my understanding into the following proof. As I'm still new to this topic, I would greatly appreciate feedback on my work.

Prove that $\lim_{x \to a} x^2 = a^2$.

Let $\epsilon > 0$ be given. We wish to show that there exists $\delta$ such that $0 < |x-a| < \delta$ implies $|x^2 - a^2| < \epsilon$.

Working backwards, note that $|x^2 - a^2| < \epsilon \iff |x-a||x+a| < \epsilon$.

If $\delta$ exists at all, there must also exist a value of $\delta \leq 1$ that works. (☛ I found this a better reason than "because $\delta$ is typically small, we suppose $\delta \leq 1,$" as that latter leaves one wondering, "What if $\delta$ happens to be large?") Thus, it is sufficient to consider if it's possible to find $\delta \leq 1$. (☛ While I'm fairly confident that this is a correct claim, I'm not entirely sure). When $|x - a| < \delta \leq 1$,

$$|x + a| = |(x -a) + 2a| \leq |x-a| + 2|a| < 1 + 2|a| \implies |x - a||x + a| < |x - a|(1+2|a|)$$

Thus, as long as we also have $|x-a|(1+2|a|) < \epsilon$ or $|x-a| < \dfrac{\epsilon}{1+2|a|}$, we will have $|x-a||x+a| < \epsilon$.

To sum up, if it's possible to find a working $\delta$, it's also possible to find a $\delta \leq 1$. For $\delta \leq 1$, we will have a working $\delta$ if it also satisfies $\delta < \dfrac{\epsilon}{1+2|a|}$.(☛ While this statement sounds good to me in its current phrasing, if I had phrased it differently, i.e. "If it's possible to find $\delta$ at all, then it's possible to find $\delta \leq 1$. If it's possible to find $\delta \leq 1,$ then $\delta$ must be less than $\dfrac{\epsilon}{1+2|a|}$." At this point, it seems a bit illogical to conclude that "thus, $\delta$ exists" as the premises have been entirely hypothetical.)

By taking $\delta = \min\left\{1, \dfrac{\epsilon}{1+2|a|}\right\}$, we will have $$|x - a| < \delta \implies |x-a| < \dfrac{\epsilon}{1+2|a|} < \dfrac{\epsilon}{|x+a|} \implies |x^2 - a^2| < \epsilon$$ It follows that $\lim_{x \to a} f(x) = a^2$.