If $p$ is an element of $\overline E$ but not a limit point of $E$, then why is there a $p' \in E$ such that $d(p, p') < \varepsilon$? I don't understand one of the steps of the proof of Theorem 3.10(a) in Baby Rudin. Here's the theorem and the proof up to where I'm stuck:
Relevant Definitions
The closure of the subset $E$ of some metric space is the union of $E$ with the set of all its limit points.
The diameter of the subset $E$ of some metric space is the supremum of the set of all pairwise distances between its elements.
For the points $x$ and $y$ in some metric space, $d(x, y)$ denotes the distance between $x$ and $y$.
Theorem 3.10(a) If $\overline{E}$ is the closure of a set $E$ in a metric space $X$, then
$$
\text{diam} \ \overline{E} = \text{diam} \ E.
$$
Proof. Since $E \subseteq \overline{E}$, we have 
$$\begin{equation*}
\text{diam} \ E \leq \text{diam} \ \overline{E}.
\end{equation*}$$
Let $\epsilon > 0$, and pick $p, q \in \overline{E}$. By the definition of $\overline{E}$, there are points $p', q' \in E$ such that $d(p,p') < \epsilon$ and $d(q, q') < \epsilon$...
I see that this works if $p$ and $q$ are limit points of $E$. But how does this work if, say, $p$ isn't a limit point of $E$? What if $E$ is some region in $\mathbb{R}^2$ along with the point $p$ by itself way off somewhere?

Basically Rudin needs to write a triangle inequality with two additional arbitrary points $p',q'$.
$d(p,q) \leq d(p,p') + d(p',q) $
$ d(p,p') + d(p',q) \leq d(p,p') + d(p',q') + d(q',q)$
He assumes $E$ is non-empty, so we might pick points $p'$ and $q'$ for this inequality, even if $p = p'$ and $q = q'$.
