If $E\subset \mathbb{R}$ is bounded then $x=\sup E$ is in $\overline{E}$ Please let me know if you think my proof is rigorous enough.
Notation: $\overline{E}$ - the closure of E; $\partial E$ - the boundary of $E$; $E^\circ$ - the interior of $E$.
If $E$ is closed then $E=\overline{E}$, which implies that for any $y\in E$, $y\le \sup E \in E$.
If $E$ is not closed then $\overline{E} = E^\circ \cup \partial E$, so that $\partial E$ contains the endpoint(s) of $E$. Hence, the endpoints of $E$ are contained in $\overline{E}$, and $x\in \overline{E}$.
 A: I'm confused by your proof, but the fact that you don't use the definition of sup is probably a sign you're not being rigorous enough.
Here's a simple proof that uses limit points:
Let $\sup E = a$ and let $n\in\mathbb{N}$. Then $\exists e_n\in E$ such that $a-e_n<\frac{1}{n}$, as otherwise $a-1/n$ would be an upper bound of $E$ that's less than $a$. Then the sequence $\{e_n\}$ converges to $a$ and so $a$ is a limit point of $E$, so it falls in the closure of $E$.
Here's the same idea but without limit points:
Let $\sup E = a$ and let $n\in\mathbb{N}$. Then $\exists e_n\in E$ such that $a-e_n<\frac{1}{n}$, as otherwise $a-1/n$ would be an upper bound of $E$ that's less than $a$. Fix $\epsilon>0$. Notice that $\exists N$ such that $\epsilon>1/N$. Thus $e_N\in B_\epsilon(x)$. But this ball clearly contains a point in $E^c$, because it contains a point larger than $x$, and $x=\sup E$.
A: For a proof to be rigorous, one necessary condition is that all the terms used in it are precisely defined (whether by you or by previous mathematicians). Now what is "endpoint" supposed to mean? I can't think of any reasonable definition that sounds like "endpoint". Like what are the endpoints of $\{ x : x \in (0,1) \land x \in \mathbb{Q} \}$? And once you fix the definition of "endpoint" you're stuck with it and have to prove whatever 'fact' you use concerning "endpoints".
A: The result follows closely from (a) the definition of closure of E.
Let sup(E) = UB and consider an open ball B(U,r) around UB.  This open set must contain points of E, for if this were not the case, we could find a value of r such that B(U,r) contains no points of E,  so we can reduce the upper bound to UB-r, so this would contradict the assumption that UB is sup(E).  Hence, any open set around sup(E) would contain points of E, this is the definition of membership of the closure of E. 
