While reading about limit points, I came upon a theorem that states that the supremum of a bounded non-empty set $S$ in $\mathbb{R}$ is a limit point of that set. In the proof, it is used that $\forall \epsilon > 0$, $\exists x_0 \in S$ such that $M-\epsilon < x_0 < M$, where $M = \sup S$. Now, I have a problem in this part. I think it should be $M-\epsilon < x_0 \leq M$ and then, the proof would not hold anymore. (Consider the set $(4,5) \cup \{6\}$ for example). Since I am using a pretty famous book, baby Rudin, it is likely that I am making a mistake. It'd be great if anyone can clarify. Here is a picture of the theorem: 
Sorry for the confusion. I misunderstood the statement. It doesn't say that the supremum is a limit point. Instead, that it's in the closure of the set. Thanks @Bungo for clarification.
Theorem 2.28 and its proof do not state nor imply that $\sup E$ is a limit point of $E.$ It states that if $E$ is closed and if $\sup E$ exists then $\sup E \in E.$ This is done by contradiction: Suppose instead that $E=\bar E,$ and $y=\sup E$ exists but $y\not \in E.$ Then,by def'n of $\sup,$ we have $(y-h,y]\cap E\ne \emptyset$ for all $h>0.$ But (because $y\not \in E$) this implies $(y-h,y)\cap E\ne \emptyset$ for all $h>0,$ implying that $y$ is a limit point of $E,$ and hence $y\in \bar E.$
In summary, $\sup E=y\not \in E=\bar E\implies (y$ is a limit point of $E)\implies y\in \bar E=E\implies y\in E,$ a contradiction.
We can also prove this directly by noting that for any $F\subset \mathbb R,$ we have (1) if $y=\sup F=\max F$ then by def'n of $\max$ we have $y\in F\subset \bar F,$ and (2) if $y=\sup F\ne \max F$ then $F\subset (-\infty, \sup F)\implies y\in \bar F$ because $F\cap (y-r,y)\ne \emptyset$ for all $r>0.$