Let $S\subseteq \mathbb R$. Show that $\sup(S') \le \sup(S)$, where $S'$ is the derived set of $S$. 
Question: If $S$ is a nonempty bounded subset of $\mathbb{R}$ then prove that $\sup(S') ≤ \sup(S)$, where $S'$ is the derived set of $S$.

$S$ is bounded. Therefore, $S ⊆[a,b]$. Let $\beta =\sup (S)$. (i.e  $x\leq \beta$ for all $x\in S$ and if $y<\beta$ then $y$ is not an upper bound for $S$)
$S'= \{x : x \text{ is a limit point of } S\}$. 
Let $\alpha = \sup (S')$. Using the property, $S\subseteq T \Rightarrow S'\subseteq T'$ I have $S'⊆[a,b]$. 
I need to show $\alpha \leq \beta$. 
 A: A great property of the supremum of a set (that you can show easily)is the following:

If $\alpha=\sup(S)$, then for every $\epsilon>0$ there is an element
  $x\in S$ such that $\alpha-\epsilon<x\leq \alpha$.

Suppose then that $\beta=\sup S<\alpha=\sup S'$. Put $\epsilon=\dfrac{\alpha-\beta}{2}>0$. By the property above, there is an element $y\in S'$ such that $\alpha-\epsilon<y<\alpha$. 
On the other hand, since $y\in S'$, $y$ is a limit point of $S$, so there exists an element $x\in S$ such that $|x-y|<\epsilon$, or equivalently, $y-\epsilon<x$. 
Draw all this elements, in a line. Can you see the contradiction?
A: 
First fact: if $A\subseteq B$, then $\sup A\le\sup B$.
Second fact: if $\bar{S}$ denotes the closure of $S$, then $\sup S=\sup\bar{S}$.
Third fact: $S'\subseteq\bar{S}$.

Proof of the first fact.  Every upper bound for $B$ is also an upper bound for $A$.
Proof of the second fact. In view of the first fact, we just need to show that $\sup\bar{S}\le\sup S$. Let $b$ be an upper bound for $S$; if $b$ is not an upper bound for $\bar{S}$, there exists $c\in\bar{S}$ such that $b<c$. By definition of closure, there is $a\in S$ such that $|c-a|<c-b$; but then $a>b$, a contradiction.
Proof of the third fact. It follows from the definition of $S'$.
Hence, from $S'\subseteq\bar{S}$ we have
$$
\sup S'\le\sup\bar{S}=\sup S
$$
