When you’re trying to show that $f$ is continuous at some $x_0$, you care only about how $f$ behaves at values of $x$ close to (but not equal to) $x_0$. Thus, if $x_0>0$, values of $x$ further away than $\dfrac{x_0}2$ don’t matter: they don’t get arbitrarily close to $x_0$, and as a result we don’t care what happens when $|x-x_0|\ge\dfrac{x_0}2$.
For a possibly simpler parallel, consider a convergent sequence $\langle x_n:n\in\Bbb N\rangle$, say with limit $L$. The convergence of the sequence to $L$ doesn’t depend on the first $100$ terms: you can change them to anything you like, and the sequence will still converge to $L$. You can change the first million terms, or the first billion, and the sequence will still converge to $L$. Convergence depends only on tails of the sequence, so as long as you leave a whole tail unchanged $-$ even if that tail doesn’t start until the $10^{10^{10}}$-th term $-$ the sequence will still converge to $L$.
Similarly, $\lim\limits_{x\to x_0}f(x)$ depends only on values of $f$ near $x_0$; you can change $f$ completely outside of any interval $(x_0-\epsilon,x_0+\epsilon)$, and you won’t change $\lim\limits_{x\to x_0}f(x)$ or $f(x_0)$. If the two were equal before the change, they’ll still be equal after the change, and $f$ will still be continuous at $x_0$.
The reasoning in the proof of the continuity of $f(x)=\dfrac1x$ on $(0,\to)$ relies on this. The first step in getting an open interval around $x_0$ in which $|f(x)-f(x_0)|<\epsilon$ is to look at the interval $\left(x_0-\frac{x_0}2,x_0+\frac{x_0}2\right)$; this isn’t small enough to do the job, but it is small enough to let us calculate a $\delta$ that will do the job.
It’s not the only possible choice. Suppose that we start by considering only values of $x$ in the interval $\left(x_0-\frac{x_0}3,x_0+\frac{x_0}3\right)$, i.e., those such that $|x-x_0|<\frac{x_0}3$. Then $\frac{2x_0}3, $x+x_0<\frac{7x_0}3$, and
$|f(x)-f(x_0)|<\frac{|x-x_0|\cdot\frac{7x_0}3}{\left(\frac{x_0}3\right)^2x_0^2}=\frac{21|x-x_0|}{x_0^3}\;.$
Thus, if you take $\delta=\min\left\{\frac{x_0}3,\frac{x_0^3\epsilon}{21}\right\}\;,$
you ensure that $|f(x)-f(x_0)|<\epsilon$ whenever $|x-x_0|<\delta$. Since $epsilon>0$ is arbitrary, and this shows how to find a satisfactory $\delta$ no matter what $\epsilon$ might be, this shows that $f$ is continuous at $x_0$. And since the computation of a satisfactory $\delta$ depends only on the fact that $x_0>0$, this in turn shows that $f$ is continuous at every positive $x_0$.