I'd appreciate some input in how I can (appropriately) simplify the reasoning in the following simple proof:
Let $E \subset \mathbb{R}$ be non-empty and bounded above. Let $y =$ sup $E$.
Prove $y$ is in the closure of $E$.
Proof: Let $\overline{E} = E \cup E'$ denote the closure of $E$, where $E'$ is the set of all limit points of $E$.
If $y \in E$, then clearly $y \in \overline{E}$.
Suppose $y \notin E$. For all $r > 0$, there must be an $x \in E$ such that $y - r < x < y$ since if it were not the case, $y - r$ would be a lower bound, contradicting $y$ being the least upper bound. Therefore we have a neighborhood centered at $y$ with an arbitrary radius $r > 0$ that contains $x \in E$. Therefore $y$ is a limit point of $E$, so $y \in E'$ and thus $y \in \overline{E}$. $\blacksquare$
The aim is to construct a sequence $(a_n)$ in $A$ such that it converges to the supremum. Then the supremum is a limit point pf $A$ and by definition of closure, it will belong to $\overline{A}$.
Proof:
Let $\operatorname{sup}(A)=a$. Since $a$ is the supremum of $A$, for $\epsilon>0$, there is an element $x\in A$ such that $a-\epsilon<x$. For the particular case that $\epsilon=1/n$ ($n$ integer), we can define a sequence $(a_n)$ in $A$ such that
$$ a - 1/n < a_n < a. $$
By the Squeeze theorem, $(a_n)\rightarrow a$. Thus, $a$ is a limit point of $A$ and $a\in\overline{A}$.