Let $(P,\le)$ be a finite poset. An element $z \in P$ is an upper bound for $x,y \in P$ if $x \le z$ and $y \le z$.
How do I prove that if every two elements in $P$ have an upper bound then $(P,\le)$ has a maximum element m such that $p \le m $ for all $p \in P$.
Also how would I show using that proof that guarantees $(P,\le)$ has a minimum element.
The claim is only true if $P$ is not empty. Here is a quick sketch for a proof: Since $P$ is not empty, let $x_1\in P$. If $x_1$ is a maximum, that's it. Otherwise, there exists some $y_1\in P$ such that $y_1\le x$ does not hold. By the given condition, there exists some $x_2\in P$ which is an upper bound for $x_1,y_1$. Argue that $x_2\ne x_1$, and thus that $x_1<x_2$. This process can be continued to find either a maximum or to construct an infinite chain $x_1<x_2<x_3<\cdots $.
The existence of minimum follows in a similar way, or, alternatively, by consider the opposite order and using the result above.