If $E$ is non-empty and bounded above, then $\sup E \in \overline E$ 
If $E \subset \mathbb{R}$ is non-empty and bounded above, then $\sup E \in \overline E$ (the closure of $E$)

This is a theorem in Rudin's real analysis (theorem 2.28,p35) but I would like to know under what circumstances it holds. The proof provided is the following:
Write $y := \sup E$, if $y \in E$, then $y \in \overline E$. If $y \notin E$, then for every $h >0$, there exists $x \in E$ such that $y-h < x < y$. Thus $y$ is a limit point of $E$ and $y \in \overline E \quad \triangle$
Now, I understand that this theorem is true for the Euclidean distance function (i.e. $d(x,y):= |y - x|)$, but does the theorem hold for all other metrics we can define on $\mathbb{R}$ as well? My guess would be no, since I was unable to prove this so I think Rudin proved this theorem thinking about the Euclidean metric.
 A: On one condition: That the new metric yields the same open and closed sets as the old one (i.e. it doesn't change the topology; it is a so-called (topologically) equivalent metric).
The reason is that the standard topology that accompanies the standard metric, is the same as the topology that comes form the standard ordering relation. By that I mean that the intervals $(a, b) = \{x\mid a<x<b\}$ are open sets, and all open sets are unions of open sets of that form (this so-called order topology may be constructed on any set with a total order relation).
If the topology induced by the new metric is not the same as the one induced by the ordering, then there is no reason to believe that the $\sup$ (which comes from the ordering) plays nicely with the closure (which comes from the metric).
A: This is actually quite a subtle question, and it's answer lies in Topology (a bit more than what Chapter 2 of Rudin covers)
The key point here is the concept of the definition of a limit point.
If you know a bit more of topology, then for any metric space $(X, d)$ with $d$ denoting the metric on $X$ a canonical topology is induced by the metric. Take $E \subseteq X$, a point $y \in E$ is a limit point of $E$ if every open ball around $B_{(E. d)}(y, h) = \{x\in E \ |\ d(y, x) < h\}$ has nonempty intersection with $E$, i.e. $B_{(E. d)}(y, h) \cap E \neq \emptyset$
The thing is that the topology and hence the open sets (the open balls) are determined by the metric $d$, there are certain conditions under which different metrics induce the same topology, but all metrics certainly do not induce the same topology. Hence in the parent space $X$ equipped with a different metric say $k$ (thus the parent space would be the topological space $(X, d)$), we have that $y$ might not be a limit point of $E \subseteq X$.
Hence when proving that any point is a limit point we always have to consider the topology, (and when we work in metric spaces we have to consider the metric), since the concept of a limit point is in fact a topological concept.
So to answer your question, no, it would not hold for all metrics, (in fact Rudin is working with the order topology as it seems in this proof). As an aside the topology induced by the usual metric and the order topology coincides on $\mathbb{R}$
A: No, the above theorem does not hold for all other metrics we can define on $\mathbb{R}$.
Take for example the discrete distance $d(x,y)$ which is zero if $x=y$ and $1$ otherwise. Then  every subset of $\mathbb{R}$ endowed with this metric is both open and closed (because every singleton is open).
Therefore if $E=(0,1)$ then $E$ is bounded above and $\sup(E)=1$ ($1$ is the least upper bound with respect to the order defined in $\mathbb{R}$, the metric is not involved here), but $1 \not\in \overline{E}=E=(0,1)$.
