Proving non-uniform continuity of functions. Say we wanted to show that $f(x)=\frac{1}{x}$ was not uniformly continuous on $(0,1)$, I will restate a proof I saw on another question, or rather a hint that I saw and my attempt to formulate a proof of it. We can strive for some contradiction, supposing that is was uniformly continuous on $(0,1)$,
$$\forall \epsilon > 0 \ \exists \delta > 0 \ \forall x,y \in (0,1) \ : \ |x-y|<\delta \implies |\frac{1}{x}-\frac{1}{y}|<\epsilon.$$
Then fix some $\epsilon=\frac{1}{2}$, and choosing $x=\frac{1}{n}$ and $y=\frac{1}{n+1}$ for some $n \in \mathbb{N}$, then we can certainly find some $\delta$ such that $|\frac{1}{n}-\frac{1}{n+1}|<\delta$, however this implies that $1<\frac{1}{2}$, which is clearly a contradiction. Then this is supposed to prove that it isn't uniformly continuous on $(0,1)$?
What I was thinking is that wouldn't this be practically identical to choosing some fixed $x=\frac{1}{3}$ and $y=\frac{1}{2}$, then we can again find a $\delta$ such that $|\frac{1}{3}-\frac{1}{2}|<\delta$ which implies that $1<\frac{1}{2}$ again.
What is the difference between the two approaches? Intuitively speaking, we would want to find some $\delta$ that works for all $x,y$, however I don't see how the choice of $\delta$ satisfying $|\frac{1}{n}-\frac{1}{n+1}|<\delta$ works for all $x,y$ as any choice of $n$ will not give you every combination of $x,y$, one because there exists rationals that cannot be written in such a form and the irrationals are also dense in the reals.
I need some verification for how these kinds of proofs of non-uniform continuity works.
 A: There is some confusion here. You chose to take $\varepsilon=\frac12$; that's fine. Then, if your function was uniformly continuous, there would be a $\delta>0$ such that, whenever $|x-y|<\delta$, then $\left|\frac1x-\frac1y\right|<\frac12$. It turns out that$$\lim_{n\to\infty}\left|\frac1n-\frac1{n+1}\right|=0,$$and therefore $\left|\frac1n-\frac1{n+1}\right|<\delta$ if $n$ is large enough. The idea is not to find a $\delta$ such that $\left|\frac1n-\frac1{n+1}\right|<\delta$; instead, the idea is to use the fact that we know that such a $\delta$ exists.
But if you only have the numbers $\frac12$ and $\frac13$, there is no way for you to be sure that $\left|\frac13-\frac12\right|<\delta$, since you don't know the value of $\delta$.
A: 
What I was thinking is that wouldn't this be practically identical to choosing some fixed $x=\frac{1}{3}$ and $y=\frac{1}{2}$, then we can again find a $\delta$ such that $|\frac{1}{3}-\frac{1}{2}|<\delta$ which implies that $1<\frac{1}{2}$ again.

And you are right. Formulated this way, this argument seems to imply more. 
But the argument is wrong, precisely here 

and choosing $x=\frac 1n$ and $y= \frac 1{n+1}$ for some $n\in\mathbb N$, then we can certainly find some $\delta$ such that $|\frac 1n − \frac 1{n+1}|<\delta$

it should read 

and for any $\delta > 0$ we can choose $x=\frac 1n$ and $y= \frac 1{n+1}$ with $n\in\mathbb N$ large enough so that $|\frac 1n − \frac 1{n+1}|<\delta$

or something like this. 

Probably you are just confused by logical quantifiers. In general, if you aim to prove a formula of the form 
$$
\forall \, \epsilon > 0 \ \exists \, \delta > 0 \ \forall \, x,y \in (0,1) \ : \ p
$$
(for example, uniform continuity of some function), you need to prove 
$$
\exists \, \delta > 0 \ \forall \, x,y \in (0,1) \ : \ p
$$
for any $\epsilon > 0$. So, given $\epsilon$ you choose $\delta$, and then $p$ is supposed hold for any $(x,y) \in (0,1)$. Unformally, you can think that your opponent chooses $\epsilon$, then you choose $\delta$, then your opponent chooses $x,y$. 

If you are proving that a given function is not uniformly continuous, you aim at 
$$
\neg \forall \, \epsilon > 0 \ \exists \, \delta > 0 \ \forall \, x,y \in (0,1) \ : \ p,
$$
or in other words, 
$$
\exists \, \epsilon > 0 \ \forall \, \delta > 0 \ \exists \, x,y \in (0,1) \ : \ \neg p.
$$
So the game is quite the opposite: you get to choose $\epsilon$, then your opponent chooses $\delta$, then you choose $x$ and $y$. If you can do it so that $p$ is always false, you win. 
