Confused about proof that diameter of a closure of a set is the same as the diameter of the set.

Note that the right hand side of the inequality $d(p, q) < 2\epsilon + \operatorname{diam}E$ is a constant independent of $p$ and $q$, so we see that $2\epsilon + \operatorname{diam}E$ is an upper bound for $\{d(p, q) \mid p, q \in \overline{E}\}$. As such, $\operatorname{diam}\overline{E}$, the least upper bound for $\{d(p, q) \mid p, q \in \overline{E}\}$, is less than or equal to this upper bound. That is, $\operatorname{diam}\overline{E} \leq 2\epsilon + \operatorname{diam}E$.

In general, if $S \subset \mathbb{R}$ and $s \leq M$ for all $s \in S$, then $\sup S \leq M$.

Hint: if $S$ is a non-empty (bounded) set of real numbers, to show $\sup S \le b$, it is sufficient to show that for any $x \in S$, $x \le b$.