($\Leftarrow$) Fix $s \in F$ such that $(\forall \epsilon)(\forall x \in A)(x < s + \epsilon)$
and $(\forall epsilon > 0)(\exists y \in A)(y > s- \epsilon)$.
Suppose for a contradiction $s$ is not an upper bound.
Then there exists $a \in A$ such that $s < a$.
Therefore, $(\forall epsilon > 0)(\forall x \in A)(x < a + \epsilon)$.
Let $\epsilon = s - a$. Then $(\forall x \in A)(x < s) \so a < s$, a contradiction.
Therefore, $s$ is an upper bound.
We claim $s$ is a least upper bound. Let $b$ be an upper bound.
Suppose for a contradiction $b < s$. We know $(\forall \epsilon > 0)(\exists y \in A)(y > s - \epsilon)$.
Let $\epsilon = s - b$. Then $(\exists y \in A)(y > b)$, contadicing that $b$ is an upper bound.
Therefore, $b \geq s \so \sup A = s$.
\section{Unions and intersections}
\subsection{Less than}
Suppose for a contradiction $\sup A > \sup B$.
We know $\sup A$ is a least upper bound.
We also know $\mathbb R$ is a complete ordered field.
Therefore, let $\sup B < x < \sup A$ for real $x$.
Because $x > \sup B$, $(\forall b \in B)(x > b) \so (\forall a \in A)(x > a)$.
Therefore $x$ is an upper bound of $A$ less than $\sup A$, a contradiction.
Therefore, $\sup A > \sup B$.
We claim that if $A \subseteq B$ and $\inf A$ and $\inf B$ exist, then $\inf A \geq \inf B$.
Suppose for a contradiction otherwise. Then let $\inf A < x < \inf B$ for some real $x$.
Because $x < \inf B$, $(\forall b \in B)(x < b) \so (\forall a \in A)(x < a)$.
Therefore, $x$ is a lower bound of $A$ greater than $\inf A$, a contradiction.