I recently was asked to show that $\frac{1}{x}$ is not uniformly continuous on $(0,\infty)$. I couldn't find a nice $x,y \in (0, \infty)$ so that I could reach a contradiciton in my proof. I have seen that people sometimes assume $\delta < 1$. I did this in the proof and was able to arrive at $\frac{1}{\delta} > \varepsilon = 1$.
My question is it valid to assume that $\delta$ is small ($\delta < 1$) and arrive at a contradiction to show that my $f$ is not uniformly continuous on $(0, \infty).$ My thinking is that if $|x-y| < \delta$, then for any $\delta_1 \geq 1$, $|x-y| < \delta_1$ should hold.
$$\forall \epsilon>0\colon\exists \delta>0\colon \forall x,y\in\Bbb R\colon \bigl(|x-y|<\delta\to |f(x)-f(y)|<\epsilon\bigr)$$ is equivalent to $$\forall \epsilon>0\colon\exists 1>\delta>0\colon \forall x,y\in\Bbb R\colon \bigl(|x-y|<\delta\to |f(x)-f(y)|<\epsilon\bigr).$$ The non-trivial direction follows because $|x-y|<\delta\to |f(x)-f(y)|<\epsilon$ gets weaker if we make $\delta$ smaller.