For any $a \ge 0$, $a < \epsilon$ for any real $\epsilon > 0$ implies $a = 0$? 
Theorem. Let ${p_n}$ be a sequence in a metric space $X$ and $p, p' \in X$. If ${p_n}\to p$ and ${p_n}\to p'$, then $p = p'$.

In the proof of this theorem, baby Rudin uses that $$d(p, p') < \epsilon$$ for arbritary $\epsilon > 0$ implies $$d(p, p') = 0.$$
I can't see why this is true because, in the real line, no matter how small you choose a $\epsilon > 0$, there will always be a $r \in \mathbb{R}$ such that $0 < r < \epsilon$, so $d(p, p')$ need not be zero for any choice of small $\epsilon$. 
Obs.: $d$ is the distance in $X$.
 A: Proof by contradiction:
If $d(p,p')$ were not $0$, then there would be a real number smaller than it.
A: The point is, $d(p, p')$ has to be some constant number, say, $d_0$, determined prior to the choice of $\epsilon$. In this way you cannot change it to an arbitrary $r$.
On the contrary, similar arguments can be used to prove that $d_0=d(p, p')=0$. If $d_0\neq 0$, there would be some $0 < r < d_0$. Letting $\epsilon = r$ yields a contradiction that $\epsilon < d_0 < \epsilon$.
A: Assume that $$(\forall \epsilon >0)\;\;d (a,b)<\epsilon $$ and let us prove that $$a=b $$.
suppose $a\neq b $ and put $\epsilon'=d(a,b)>0 $ then
$$d (a,b)<\epsilon'$$
$$\implies d (a,b)<d (a,b) $$
which has nosense.

or

$$(\forall \epsilon>0)\;\;0\leq d (a,b)<\epsilon$$
$$\implies $$
$$(\forall n\in \mathbb N)\;\;\;0\leq d (a,b)<\frac {1}{n+1} $$
$$\implies $$
$$0\leq d (a,b)\leq \lim_{n\to+\infty}\frac{1}{ n+1}$$
$$\implies 0\leq d (a,b)\leq 0$$
$$\implies d (a,b)=0\implies a=b $$
A: "$d(p,p') < \epsilon$ for arbitrary $\epsilon > 0$" is another way of saying this:

$d(p,p')$ is strictly less than every positive number.

Since $d$ is a metric, then $d(x,y) \ge 0$ for all $x,y \in X$.  Therefore:

$d(p,p') \ge 0$ by definition, and $d(p,p')$ is strictly less than every positive number.

There's only one number that satisfies both of those conditions.
A: Notice that $\epsilon$ is arbitrarily small, so in passage to a limit we can take this distance to be zero.
