The set $S = \{1/n:n\in\Bbb{N}\}$ is not open, because for example it contains the point $1$ but it does not contain any $\varepsilon$-neighbourhood...
See my comment on your previous thread. The way this proof works in that you divide the real line into subintervals of length $1/2^n$ and you locate...
No. What is happening in this proof is that you narrow down the location of $\sup(S)$ by an approximation process. You start from a point $s$ in $S$,...