I have seen that this question has been answered already, but I wanted to come up with my own proof and would appreciate some feedback, as I am not always the best at transferring what is in my head into words and math symbology.
Proof: By definition, $\overline{A} = A \cup L$, where $L$ is the set of all limit points of $A$. In order for $s \in \overline{A}$, either $s \in A$ or $s \in L$.
Suppose $s \in A$. Then we are done, as $s \in A \Rightarrow s \in A \cup L \Rightarrow s \in \overline{A}$.
Now suppose that $s \notin A$. By definition, as $s=$ sup $A$, $s$ is the least upper bound of $A$. Let $V_\varepsilon(s)$ be an $\varepsilon$-neighborhood of $s$. Then no matter how small we make $\varepsilon$, $V_\varepsilon(s) \cap A \neq \emptyset$ since $s=$ sup $A$, which implies that $s \in L$. Then $s \in L \Rightarrow s\in A \cup L \Rightarrow s\in \overline{A}$.
Q.E.D.
Even if this proof is correct, I still don't feel great about the line where I write "Then, no matter how small we make $\varepsilon$..." I guess it just doesn't feel very mathy to me, and I am also not sure it transitions as well as it could from the idea that $s=$ sup $A$.
I think "no matter how small we make $\epsilon$" is how I would describe it if in conversation, but I can definitely get the desire to make it more 'mathy', particulary if it's for a graded assignment.
If you want to 'math it up a bit', my suggestion is that you spend a couple lines proving why $V_\epsilon(s)\cap A\neq\emptyset$ for all $\epsilon > 0$, possibly with a contradiction proof. For instance, if there is an $\epsilon >0$ such that $V_\epsilon(s)\cap A =\emptyset$, what can you say about some of the elements in this neighborhood and their relation to both $s$ and to all of the elements of $A$?