$\varepsilon$-$\delta$-proof$:$ $f(x)=\frac{1}{x}$ is continuous in $[1, \infty)$

With $\varepsilon$-$\delta$ prove that $f(x)=\frac{1}{x}$ is continuous in $[1, \infty)$.

Let $\varepsilon > 0$, let $\delta = \varepsilon$ and if $|x-x_{0}|< \delta$ then

$$\left|\frac{1}{x}-\frac{1}{x_{0}}\right|= \left| \frac{x_{0}-x}{x \cdot x_{0}} \right| < \left|x_{0}-x\right|=$$

$$= |-(-x_{0}+x)|=|-(x-x_{0})|< |-(\delta)|< |-\delta|=\varepsilon$$

Please let me know if everything is alright and be as strict as possible (maybe not too strict because I'm visiting a beginners course of analysis!). But strict as in an exam which I'll write in a few days... :o

I'm especially not sure if the first "<" I used is allowed and how to deal with $|-\delta|$ (I would say the negative sign will be ignored because of factorial and $[1, \infty)$?)

I'd also like to know other ways of proofing continuity for this function if possible!

Looks correct, except the first $<$ should be $\le$ (what if $x = x_0 = 1$?). You should probably explain that step (say something like "because $|xx_0| \ge 1$").
And it could be much shorter: $$\left|\frac{1}{x}-\frac{1}{x_{0}}\right|= \left| \frac{x_{0}-x}{xx_{0}} \right| \le |x_{0}-x| = |x - x_0| < \delta = \epsilon.$$