Let $A$ denote a closed subset of $\mathbb{R}.$
It seems obvious that for all $B \subseteq A$ we have that if $B$ has a supremum, then $\mathrm{sup}(B) \in A$. Perhaps this is trivial, but I cannot see how to show it. Ideas, anyone? I am especially interested in a proof that makes no mention of the metric space structure on $\mathbb{R}$. A purely topological/order theoretic proof.
Suppose $\emptyset\ne B\subseteq A\subseteq\mathbb R,\ $ $A$ is closed, and $\sup B=s\in\mathbb R$. For any $c\lt s$ we have $B\cap(c,s]\ne\emptyset$, otherwise $c$ would be an upper bound for $B$, contradicting our assumption that $s$ is the least upper bound. Hence $s\in\text{cl}(B)\subseteq\text{cl}(A)=A$.
To elaborate on the last sentence: First, $\text{cl}(B)$ is the closure of $B$, defined as the set of all points $x$ such that every neighborhood of $x$ contains a point of $B$. Any neighborhood of a real number $s$ contains an open interval of the form $(c,d)$ where $c\lt s\lt d$, which contains the interval $(c,s]$, which contains a point of $B$. Next, $\text{cl}(B)\subseteq\text{cl}(A)$ because $B\subseteq A$; the closure operator is monotone. Finally, $\text{cl}(A)=A$ because $A$ is closed; that's the definition of a closed set.