Rudin Theorem 4.20 Could you tell me how $\lvert f(t) - f(x)\rvert$ becomes greater than $\epsilon$?
In this proof, I think the author didn't even define $\epsilon$(that is, $\epsilon$ is < something). Did he?  

 A: I think the gist here is: since $\,x_0\,$ is a limit point of $\,E\,$ , we can get $\,x-x_0\,$ as small as we want, which means $\,|f(x)|=\left|\frac1{x-x_0}\right|\;$ can be made arbitrarily big , and:
$$\left|f(t)-f(x)\right|=\left|\frac1{t-x_0}-\frac1{x-x_0}\right|=\left|\frac{x-t}{(t-x_0)(x-x_0)}\right|$$
so you see we can bound by delta the numerator, but not so the denominator
A: It says:

let $\varepsilon>0$ [...] be arbitrary

so the argument that follows must (and does) work no matter what $\varepsilon$ is, as long as it's positive.
$|f(t)-f(x)|$ can become greater than any desired $\varepsilon$ because at this point in the proof $f(x)$ is just a constant, and $|f(t)|$ can become as large as you want it (and therefore as far from $f(x)$ as you want) by choosing $t$ close enough to $x_0$.
A: The idea here is that no matter what $\epsilon$ or $\delta$ we pick, we can always make the value $\lvert f(x)-f(t) \rvert>\epsilon$, even for $\lvert x-t \rvert < \delta$. Restated, you can't pick an $\epsilon,\delta$ such that varying $t$ within $\delta$ of $x$ keeps the value $\lvert f(x)-f(t)\rvert$ smaller than $\epsilon$, so $f$ is not uniformly continuous.
A: Part I
Since $ x_0 $ is a limit point of $ E $ we can choose $ t \in E $ such that $ | x_0 - t | $ is as small as we want, but still greater than $ 0 $.
So we can make $ \frac{1}{| x_0 - t |} $ as big as possible, in particular:
$$ \frac{1}{| x_0 - t |} > \epsilon + \frac{1}{| x_0 - x |} $$
since $ x \in E $ was chosen in previous step (choice of $ t $ depends on choice of $ x $). But then:
$$ \frac{1}{| x_0 - t |} - \frac{1}{| x_0 - x |} > \epsilon $$
so:
$$ \frac{|x_0 - x| - |x_0 - t|}{| x_0 - x || x_0 - t |} > \epsilon $$
but since $ | x_0 - x| \leq | x - t | + | t - x_0 | $ and thus $ | x_0 - x| -| t - x_0 | \leq | x - t | $
we get:
$$ \frac{|x - t |}{| x_0 - x || x_0 - t |} \geq \frac{|x_0 - x| - |x_0 - y|}{| x_0 - x || x_0 - t |} > \epsilon $$
This shows we can make $ | f(t) - f(x) | > \epsilon $.
Part II
Note also that since $ x $ was chosen so that $ | x_0 - x | < \delta $, we can make $ | x_0 - t | $ so small so that:
$$ | x_0 - t | < \delta - | x_0 - x | $$
but then:
$$ | x - t | \leq | x - x_0 | + | x_0 - t | < \delta $$
Conclusion
So overall we can choose $ t $ which fulfills both conditions.
