The Encyclopedia of Mathematics claims
In a lattice satisfying the descending chain condition (...) the join-irreducible elements are those which cover precisely one element...
I have been unable to prove this statement, any tips or a complete proof of this would be appreciated.
There is a mistake in the article's claim: it should be ascending chain condition, not descending (or alternatively "join-irreducible" and "cover" should be replaced by "meet-irreducible" and "is covered by" respectively). For an easy counterexample, consider $\omega+1$ as a lattice; the top element does not cover anything, but is join-irreducible.
With that out of the way, let me prove the correct claim. Some terminology for readers:
An element $x$ of a lattice $L$ is join-irreducible iff there is some $y<x$ and $x$ is not the join of a finite antichain of size $>1$.
$x$ is a cover of $y$ iff $x>y$ and there is no $z$ with $x>z>y$.
It is easy to see that if $x$ is join-irreducible then $x$ cannot cover more than one element: if $x$ covers both $y$ and $z$, then $x=y\vee z$ but $\{y,z\}$ is a finite antichain with more than one element. (Note that ACC is not used here.)
So we just need to show that if $x$ is join-irreducible and $L$ is ACC then $x$ does cover something. This is quick: if $x$ doesn't cover anything and there is some $y_1<x$ in the first place, then we can recursively build a sequence $y_1<y_2<y_3<...<x$, contradicting ACC. (Having found $y_n$ we must be able to find $y_{n+1}\in (y_n,x)$ since $x$ doesn't cover $y_n$.)