We wish to prove that $f(x)=\frac{1}{x}$ defined on $[1/2,1]$ is uniformly continuous.
Scratch work: Take $x,y \in [1/2,1]$, $|f(y)-f(x)|=|\frac{1}{y}-\frac{1}{x}|=|\frac{y-x}{xy}|$
At most $x=1,y=1$ so at most $|xy|=1$. Then $$\frac{|y-x|}{|xy|}\le|y-x|<\delta$$ So if we choose $\delta = \varepsilon$ our proof will be complete.
Proof:
$\varepsilon>0 $ is given, choose $\delta = \varepsilon$,
$x,y\in[1/2,1]$ satisfying $|y-x|<\delta$ are given.
$|f(y)-f(x)|=|\frac{1}{y}-\frac{1}{x}|=|\frac{y-x}{xy}|\le|y-x| <\delta=\varepsilon$.
The proof is complete.
Is this all correct?