I want to show that $f(x) = x/(1+x)$ is uniform continuous on $[0, \infty)$.
My proof: Let $x,y$ be real numbers in $[0, \infty)$ and let $\epsilon$ be a positive number.
We choose $\delta = \epsilon$ such that $|y-x| < \delta = \epsilon$.
So, we have
$$ |f(x) - f(y)|\,\, =\,\, |y/(1+y) - x/(1+x)|\,\, =\,\, \left|\frac{y-x}{(1+y)\cdot(1+x)}\right|. $$
We note $(1+y)(1+x) \ge 1$, so $$|f(x) - f(y)| < |y - x| < \epsilon.$$ Thus, $f(x)$ is uniform continuous on $[0, \infty)$.
Is this proof correct?