Can someone please verify my proof or offer suggestions for improvement? There may be answers to the same questions elsewhere, but I need help with my proof in particular.
Show that if $P$ is a finite poset such that
(1) Every $x, y$ has a greatest lower bound
(2) $P$ has a unique maximal element $\hat 1$
then $P$ is a lattice.
We need to prove that every $x, y$ in $P$ has a least upper bound. Suppose, for the sake of contradiction, that there exists an $x, y \in P$ that does not have a least upper bound. There are two possible cases:
(1) $x, y$ has no upper bound: In this case, $x$ and $y$ are both maximal elements (note that it must be the case that $x \neq y$, or else $x$ would be a least upper bound for $x, y$.) This contradicts the fact that there exists a unique maximal element.
(2) $x, y$ has an upper bound $z_1$, but no least upper bound: In this case, the set of all upper bounds of $x, y$ is infinite, contradicting the fact that $P$ is a finite poset; if we are given an upper bound $z_i$, we can find a smaller upper bound $z_{i+1}$ such that $z_{i+1} \neq x, y$.
Therefore, it must be the case that every $x, y \in P$ has a least upper bound. Hence, $P$ must be a lattice.