I managed to show (a), but I am still stuck at (b). I know that in a finite lattice, every set has a minimal and maximal element.
I was thinking of maybe characterising the join irreducible elements as follows: all the sets that have an infimum and contain their infimum. Say we have a set $A$ with infimum $a$, and a set $B$ with infimum $b$. If we consider their union, then we know that this union contains a minimal element, because it is a finite subset of a lattice. How can I show that this minimal element should be $a$ or $b$?
I would also like to show that if a set has multiple minimal elements, it cannot be join irreducible, but I’m not sure how.
Could someone help me out?
Define $\uparrow\!x = \{p \in P : x \leq p\}$. Clearly, $\uparrow\!x \in\mathrm{Up}(P)$.
If $\uparrow\!x = A \cup B$, with $A, B \in \mathrm{Up}(P)$, then, since $x \in \uparrow\!x$, it follows that $x \in A$ or $x \in B$, whence $\uparrow\!x \subseteq A$ or $\uparrow\!x \subseteq B$, yielding $\uparrow\!x=A$ or $\uparrow\!x=B$.
Therefore $\uparrow\!x$ is join-irreducible in $\mathrm{Up}(P)$.
For the converse, let $A \neq \uparrow\!x$, so that $A$ has more than one minimal element. Say the minimal elements of $A$ are $a_1, \ldots, a_n$.
Then $A = \bigcup_{i=1}^n \uparrow\!a_i$, and since $i \neq j$ implies $\uparrow\!a_i \nsubseteq \uparrow\!a_j$, it follows that $A$ is not join-irreducible.
Therefore, the join-irreducible elements of $\mathrm{Up}(P)$, where $P$ is a finite poset, are precisely the sets $\uparrow\!x$ with $x \in P$.