Uniform continuity of function

Let $\epsilon > 0$ and $g : (a,\infty) \to \mathbb R$ given by $g(x) = 1/x$. Take $\delta = a^2\epsilon$. Note that for any $x,y \in (a,\infty)$, $xy \ge a^2$ so $1/(xy) \le 1/a^2$. Then if $|x-y| < \delta$,

$$ \left| \frac{1}{x} - \frac{1}{y} \right| = \frac{|y-x|}{xy} \le \frac{|y-x|}{a^2} < \frac{\delta}{a^2} = \epsilon $$

If we consider the interval $I=(a,\infty)$, then we see that, for any $x,x' \in I$: \begin{align*} \lvert f(x) - f(x') \rvert &= \left\lvert \frac{1}{x} - \frac{1}{x'} \right\rvert \\ &= \left\lvert \frac{x'-x}{x\cdot x'} \right\rvert \\ &< \frac{1}{a^2} \cdot \left\lvert x'-x \right\rvert \\ \end{align*} For any $\epsilon > 0$, we can choose $\delta = a^2\epsilon$ and see that: $$ \lvert x - x' \rvert < \delta \implies \lvert f(x) - f(x') \rvert < \epsilon \qquad \forall x,x' \in I$$ By definition, $f(x)$ is uniformly continuous on $I$.

To answer your question about why concerning the $\delta$ - the $\delta$ is not unique. Choosing a smaller $\delta$ works. Uniform continuity means that we can choose a $\delta$ that works regardless of which points are chosen - dependent only on $\epsilon$.