Why is it if $\epsilon > 0$ is arbitrary implies that $d(p,p') = 0$ if $d(p,p') < \epsilon$? I know this is such a basic question and its embarrassing to ask but I just don't understand why Rudin concludes such a thing. In theorem 3.2 (a) he proves that if two sequences converge to different limits then the limits are actually the same as follow:
$$ \epsilon > 0 $$
$$ n \geq N \implies d(p_n,p) < \frac{\epsilon}{2}$$
$$ n \geq N' \implies d(p_n,p') < \frac{\epsilon}{2}$$
let $n = \max(N,N')$:
$$ d(p,p') \leq d(p,p_n) + d(p_n,p') < \epsilon $$
everything so far made sense. The last step of the proof is what I don't understand:

since $\epsilon$ arbitrary, we conclude that $d(p,p') = 0$

thats the part that I don't understand. The proof started assuming that $ \epsilon > 0$ so of course the proof didn't conclude by plugging in zero. In chapter 2 there was such a big emphasis that the points in the neighborhoods to the limit points are different from the limit point. See this proof I'd conclude there is always a positive difference between p and p' but its never exactly zero (just like in chapter 2 and limit points). 
Of course I assume I am wrong and there some subtle but important point that I don't understand and wanted to clarify it. Can someone clarify why my assertion that the difference is always positive (of course by assumption) is wrong? If anything I'd conclude that $p$ and $p'$ are limits of each other but not that they are the same as Rudin concludes or that their distance is zero. What did I miss?

note: I do know limit points and limits aren't the same, was just saying that to provide context for my confusion.
 A: Others explained this but basically you have 
$$0\leq d(p,p')<\epsilon \ \forall \epsilon>0$$
If you let $d(p,p')>0$ then just pick $\epsilon = d(p,p')/2$ to get a contradiction. Hence the only value $d(p,p')$ can take on is zero.
A: Another way of putting it and this is probably more likely  in Rudin's thinking:
$\mathbb R$ has the least upper bound property and the basic definition of "bounded below", "lower bound" and "greatest lower bound" make the following statement almost a tautology.
Let $S\subset \mathbb R$ be a non-empty set that is bounded below.  And let $K < s$ for any $s \in S$.   Then that would mean $K \le \inf S$.  (Because if $K < s$ for all $s\in S$ then $K$ is a lower bound of $S$ so $K \le \inf S$, the greatest lower bound. [We do require that the least upper bound property holds on $\mathbb R$ however.])
Now let $S= \mathbb R^+ = \{x \in \mathbb R| x > 0\}$.  $\mathbb R^+$ is bounded below by any non-positive number.  So $\inf \mathbb R^+$ exists and it takes little effort to discover $\inf \mathbb R^+ = 0$.
$K < \epsilon$ for any $\epsilon \in \mathbb R^+$ means $K \le \inf \mathbb R^+ = 0$.
If we are further told $K \ge 0$ the we have $K \le 0$ and $K \ge 0$.  That means..... $K = 0$.
.....
But that's a lot of sound and fury for what should be intuitive once you think about it:
...
Consider the statement.  "Let $K$ be a non-negative real number that is smaller than any arbitrary positive number".  
It's subtle but that statement is possible but only if $K = 0$.
If such a number exists it can't be negative because we were told $K$ is not negative.
It can't be positive because it is smaller than any positive number so if it were positive it would be smaller than itself, which is a logic impossibility-- no number can be smaller than itself.
The only third option is $K = 0$ and that is not inconsistent.  $0$ is non-negative, and $0 < \epsilon$ for all $\epsilon > 0$.
...
A: I think I get it now. Since
$$ d(p,p') < \epsilon $$
(and $ 0 \leq d(p,p')$)
for all positive numbers then by trichotomy (i.e. one has to hold, <,>,=) then it must mean that its not equal to any positive or negative number then it must be zero (because we know all the things is not and since 1 thing does have to hold it must be equality to zero).

"I think the simplest answer would be that 0 is the only non-negative
  real number that is less than every positive real number."


On a related statement I noticed that if we change things to less than or equal that $d(p,p')=0$ is still zero. 
$$ d(p,p') \leq \epsilon $$
Proof:
Assume $\forall \epsilon > 0, d(p,p') \leq \epsilon$ AND $d(p,p') \neq 0$. Then there must exist some $\delta >0$ s.t. $d(p,p') = \delta$. Then the $d(p,p') \leq \epsilon$ can't hold for every epsilon because it doesn't hold for $\frac{ \delta}{2}$. QED.
