I looked that this question that was posted regarding this proof. I mostly understand the answer, but it states that it is required to choose $\delta$ from two different deltas that have been found in advance.
The short version is restricting the interval around $x_0$ such that $\frac{1}{2}x_0 < x < \frac{3}{2}x_0$ via a $\delta_1 = \frac{1}{2}x_0$ one arrives at $|f(x_0) - f(x)| < \frac{10}{x_0^3}|x_0 - x|$.
Thus if $\delta = \frac{x_0^3}{10}\epsilon$ we would have
$$|f(x_0) - f(x)| < \frac{10}{x_0^3}|x_0 - x|\leq \frac{10}{x_0^3}\frac{x_0^3}{10}\epsilon = \epsilon $$
Summarising:
$$|x - x_0| < \frac{1}{2}x_0 \implies \frac{|x_0 + x|}{x_0^2x^2} < \frac{10}{x_0^3} \implies |f(x_0) - f(x)| < \frac{10}{x_0^3}|x_0 - x|$$ and thus $\delta = \frac{x_0^3}{10}\epsilon$ already suffices?
Why does the answer state it is necessary to choose the minimum from the first delta and the second delta?
You choose $\delta_1 = \frac{1}{2}x_0$ and $\delta_2 = \frac{x_0^3}{10}\epsilon$.
If you let your general $\delta=\delta_2$, and if $\delta=\delta_2>\delta_1$ then the proof doesn't work because there will be $\alpha$ such that $\delta_1<\alpha<\delta$ and proof won't work for that $\alpha$. Since you don't know which delta is smaller, you choose the minimum just to guarantee your proof works.