Does this proof look fine or contain gaps? Do you have suggestions? Many thanks for your dedicated help!
Let $(A,\le)$ be an ordered set whose every nonempty subset $X$ bounded from above has a supremum. Then every nonempty subset $Y\subseteq A$ bounded from below has an infimum.
Let $Z$ be the set of lower bounds of $Y$. It's clear that $Z$ is bounded above by every elements in Y. Thus $Z$ has a supremum, which is denoted by $\sup Z$. Next we prove $\sup Z$ is the infimum of $Y$.
$\sup Z$ is the least upper bound of $Z$, and $Y$ consists of upper bounds of $Z \implies$ $\sup Z\le y$ for all $y\in Y$.
If $z$ is a lower bound of $Y$, then $z\in Z$ and consequently $z\le \sup Z$.
Thus $\sup Z$ is the infimum of $Y$.
Update: Since $Y$ is bounded from below, $Z$ is nonempty.