Show that $f(x) = \frac{1}{x}$ is uniformly continuous on $[1,\infty)$ Verification of argument. Show that $f(x) = \frac{1}{x}$ is uniformly continuous on $[1,\infty)$
Generally speaking I understand why this function is uniformly continuous, but I wanted to just verify that my logical thinking on the matter is correct.

Definition: A function $f: S \to \mathbb{R}$ is uniformly continuous if for all $\epsilon >0$, there exists $\delta >0$ such that for all $x,y \in S$ if $|x-y| < \delta \Rightarrow |f(x) - f(y)| < \epsilon$.

Analyzing the expression: $$|f(x) -f(y)| = \frac{|x-y|}{|x||y|} < \frac{\delta}{|x||y|}$$
We are given that $x,y \in [1,\infty)$ and we want $\frac{|x-y|}{|x||y|} < \epsilon$. This means that for any value in the set, $$\frac{\delta}{|x||y|} < \delta $$. So if we let $\delta = \epsilon$ then we will have a relationship such that:
$$\frac{|x-y|}{|x||y|} < \frac{\delta}{|x||y|} = \frac{\epsilon}{|x||y|} < \epsilon$$. And we are done.
I feel perhaps I could word the explanation better as to how we make the jump to concluding that the $\delta$ we chose is the right one. Feedback on the proof? THanks in advance.

You  are on the right track. 
Your $\delta$ is correct.
Just note that since $x,y\geq 1$ then $\frac{1}{xy} \leq 1$
Also another proof: this functions has bounded derivative in $[1,+\infty)$ so it is Lipschitz continuous,thus uniformly continuous.
