Consider $A\subset \mathbb{R}$, where $A$ is bounded. Since $A$ is bounded, $\sup(A)$ exists. To show that $\sup(A)\in Closure(A)$, we need to show that $\sup(A)$ is a limit point of $A$. Let $\sup(A)=s\in\mathbb{R}$. If $A$ is a finite set then it is closed, so the closure of $A$ is $A$, and $\sup(A)\in Closure(A)$. If $A$ is not finite, consider the sequence $\{s-1/n\}\cap A$. This sequence converges to $s$ and is entirely contained in $A$.
But I have some doubts about my proof approach because if we consider $A:=(0,1/2)\cup\{5\}$, the above sequence will not intersect $A$. In this case, however, one can argue that $\{5\}$ is a singleton, which is closed, and so $\{5\}$ must be in the closure of $A$. In all other cases, it seems, the above sequence will be contained in $A$ and will converge to $s$. Is this true?
Another approach I could use is this. Suppose $s\not\in Closure(A)$. Then $s\in \mathbb{R}\backslash Closure(A)$. Since $Closure(A)$ is closed, $\mathbb{R}\backslash Closure(A)$ is open, thus $\exists$ an $r>0$ such that the open ball $B_r(s)\not\cap Closure(A)$. Which is impossible if $s$ is not a singleton. If $s$ is a singleton, then $s\in Closure(A)$ by definition. Therefore, $s\in Closure(A)$ in all cases.
Please let me know about the two of my approaches.
Best Answer
To show that a real number $x$ lies in the closure $\bar{A}$ of $A$, you have to show that either $x\in A$ or that $x$ is a limit point of $A$. While some points in $A$ are limit points of $A$, your example $A=(0,\frac{1}{2})\cup\{5\}$ demonstrates that not every element of $\bar{A}$ is a limit point of $A$, as it could be an isolated point in $A$.
So to show that $\alpha=\sup(A)\in\bar{A}$, we have to show that either $\alpha\in A$ or that $\alpha$ is a limit point of $A$. If $\alpha\in A$ then we're done, so suppose that $\alpha\not\in A$.
Since $\alpha-1$ is not a upper bound for $A$, it follows that we can choose some $s_1\in A$ such that $\alpha\geq s_1>\alpha-1$. Now $s_1\neq \alpha$ because $\alpha\not\in A$, so since neither $\alpha-\frac{1}{2}$ nor $s_1$ is an upper bound for $A$, we can choose $s_2\in A$ such that $\alpha \geq s_2>\max\{s_1,\alpha-\frac{1}{2}\}$.
Proceeding inductively, we can construct a sequence $\{s_n\}$ such that $\alpha\geq s_n>\max\{s_{n-1},\alpha-\frac{1}{n}\}$ for all $n\geq 2$. This means that $\{s_n\}$ is a sequence of distinct elements of $A$ which converges to $\alpha$, hence $\alpha$ is a limit point of $A$.