Definition. Let $E$ be a nonempty subset of $X$, and let $S$ be the set of all real numbers of the form $d(p, q)$, with $p,q\in E$. The sup of $S$ is called the diameter of $E$.
Theorem 3.10. If $\overline{E}$ is the closure of a set $E$ in a metric space $X$, then $$\text{diam }\overline{E} = \text{diam }E.$$
Proof: Fix $\varepsilon>0$, and choose $p, q \in \overline{E}$. By the definition of $\overline{E}$, there are points $p',q' \in E$ such that $d(p,p') < \varepsilon$ and $d(q,q') < \varepsilon$. Hence $$d(p, q) \le d(p,p') + d(p', q') + d(q', q) < 2\varepsilon + d(p', q') \le 2\varepsilon + \text{diam }E.$$
Then, it follows that $$\text{diam }\overline{E} \le 2\varepsilon + \text{diam }E$$ Since $\epsilon$ was chosen arbitrarily, we have that $$\text{diam }\overline{E} \leq \text{diam }E.$$
My question: How does the arbitrariness of $\epsilon$ imply $\text{diam }\overline{E} \leq \text{diam }E?$ After all, $\epsilon$ can never be exactly equal to $0$. I tried a proof by contradiction for the same, to no avail.
If you have two real numbers $x$ and $y$, then $$ x\le y \iff \forall r >0 (x\le y+r) $$ The implication $\Rightarrow$ is obvious. To see the converse, suppose $\forall r>0 (x\le y+r)$ but it is not the case that $x\le y$. This means that $y<x$, so $x-y>0$ and we can pick $r$ with $0<r<x-y$. By assumption, $$ x\le y+r <y+(x-y)=x $$ This is impossible.