$f(x) = \frac{2}{x}$ is not uniformly continuous on $(0,1]$ Show that $f(x) = \frac{2}{x}$ is not uniformly continuous on $(0,1]$
WLOG Suppose, $0< \delta \leq 1.$ Let, $\epsilon = 1$ and $x = \frac{\delta}{2}, y = x + \frac{\delta}{3},  x,y \in (0,1]$
Skipping all the details: $\mid f(x) - f(y)\mid =...= \mid \frac{24}{5\delta}\mid \geq 1 = \epsilon$, since $0<\delta\leq 1$.
I have seen people restricting the $\delta$ like I did here, but I have never done this before and I thought the only condition applies to $\delta$ is $\delta >0$(though restricting the $\delta$ does make things easier). So, since this is one of my HW question, am I allowed to do it like this (just to make sure I don't make any stupid mistake)? Do you think my approach is correct? Please let me know. Thanks. 

I am really unsure that this is right, I think it's wrong but that's what I tried :Consider $x_{n}=\frac{1}{n}$, $y_{n}=\frac{1}{2n}$. $x_{n}-y_{n}=\frac{1}{2n}$. $|f(x_{n})-f(y_{n})|=2n$. Let $\epsilon =1$. Let $\delta>0$. Pick $n$ such that $\frac{1}{n}<\delta $. $|x_{n}-y_{n}|<\frac{1}{n}<\delta$ and $|f(x_{n})-f(y_{n}|>1$
