Take any = 1. Does there exist any such that
if | t - s | < then | f(t) - f(s) | < = 1 ?
The basic idea is easy: since | f(t) - f(s) | = | 1/t - 1/s | = | s - t | * 1 / | st |, we can see that this might approach infinity if s and t approach zero, and therefore will be bigger than any chosen . All we have to do now is formalize this proof.
Assume there exists such a . Without loss of generality we may assume that < 1 (why ?). Then let s = t + / 2 and set t = / 2. We have (assuming that s, t are positive):
| f(t) - f(s) | = | 1/s - 1/t | = | (t - s) / st | = > 1
But now, no matter what < 1 is we can make | f(t) - f(s) | > 1. Therefore, the function is not uniformly continuous.
This proof, loosely speaking, depends on the fact that after simplification | f(t) - f(s) | goes to infinity if s and t approach zero. That is exactly the situation as described in the above picture.