Let $E\subset\mathbf R$, and suppose that $\sup E$ exists in $\mathbf R$. It is not too difficult to prove that if $\sup E\not\in E$, then $\sup E$ is a limit point of $E$:
Let $\alpha=\sup E$, and suppose that $\alpha$ is not a limit point of $E$. Then there is a $h>0$ such that $(\alpha-h,\alpha+h)$ does not contain any members of $E$. Hence every member of $E$ must be less than or equal to $\alpha-h$. But this contradicts the fact that $\alpha$ is the least upper bound of $E$.
My question is: can this proof be generalised? More explicitly, suppose that $X$ is both an ordered set and a metric space, and let $E\subset X$. Assuming that $\sup E$ exists in $X$, must it be the case that if $\sup E\not\in E$, then $\sup E$ is a limit point of $E$? In case the answer is negative, is there an extra condition which does guarantee this property?
The answer is quite trivially negative when the order and metric are unrelated, but it is also negative in a suitably good setup, like subspace of order topology.
Consider $X=[0,1)\cup \{2\}$ and $E=[0,1)$ with the Euclidean metric and order. Note that $2$ is the supremum of $E$ but it is not its limit point. This space however is only a subspace of order topology. It is not the order topology for the inherited order (that would be homeomorphic to $[0,1]$ interval).
In general it is true for order topology over total order. In that situation let $s=\sup E\not\in E$. Then for any $x\in E$ there is $y\in E$ such that $x<y<s$. Therefore every open neighbourhood of $s$ contains some element from $E$ and so it is a limit point of $E$.
If the partial order is not total then again there are counterexamples, even for order topology. For example consider $X=\{0,1,2\}$ with $0<2$ and $1<2$, while $0$ and $1$ are unrelated. The induced order topology has only two (nontrivial) open subsets: $\{0,1\}$ and $\{2\}$. So $E=\{0,1\}$ has $2$ as the supremum but it is not its limit point.