I have been trying to understand the proof for the fact that the supremum of a closed set of real numbers belongs the set itself, as given by Theorem 2.28 in "Principles of mathematical analysis" by Walter rudin. The proof in the textbook goes as follows:
Assume the supremum y $\notin$ E. For every $h>0$ there is a point $x \in E$ such that $y-h \leq x \leq y$, since y is the supremum. I was able to understand up until this point.
Then the argument is that every neighborhood of y contains a point $x \in E$ and $x \neq y$ since $y \notin E$. Thus y is a limit point of E but $\notin E$, thus contradictory to the assumption that E is closed. My rough interpretation of the statement is that since every neighbourhood of y is non empty and any given element of the neighbourhood can be written as $y-h$ for some $h>0$, it should contain an element $x \in E$, thus making it the limit point.
Now, I am confused if this would be valid for any metric. In other words is $y-h \leq x \leq y$ equivalent to saying $d(y,y-h) < r \implies d(y,x) < r$ for any metric?