Assuming the ultrafilter lemma, compactness is equivalent to every net having a cluster point, every net having a convergent subnet, every filter having a cluster point, and every ultrafilter converging

Suppose $X$ is compact and $\mathcal F$ is a filter. The closed family $\{\overline A:A\in\mathcal F\}$ has the finite-intersection property, because a finite intersection of members of $\mathcal F$ is nonempty and is contained in the corresponding intersection of closures. By [L1], choose $p\in\bigcap_{A\in\mathcal F}\overline A$.

If every filter has a cluster point, apply this to a net's tail filter and use [L3]; hence 4 implies 2. By [L2], conditions 2 and 3 are equivalent.

Conversely, if every net has a cluster point and $\mathcal F$ is a filter, its derived net has a cluster point, which is also a cluster point of $\mathcal F$ by [L3]. Hence 2 implies 4.

Condition 4 implies 5 because an ultrafilter is a filter and [L4] turns its cluster point into a limit.

Suppose every ultrafilter converges and let $\mathcal C$ be a family of closed subsets of $X$ with the finite-intersection property. Clause 2 of [L1] gives a filter containing $\mathcal C$, and [L4] extends it to an ultrafilter $\mathcal U$.

Every neighbourhood of $p$ meets every $A\in\mathcal F$, since $p\in\overline A$; thus $p$ is a cluster point of $\mathcal F$. Hence 1 implies 4.

Let $p$ be a limit of $\mathcal U$. For $C\in\mathcal C$, every neighbourhood of $p$ belongs to $\mathcal U$ and meets $C\in\mathcal U$; therefore $p\in\overline C=C$. Thus $\bigcap\mathcal C\ne\varnothing$, and [L1] gives compactness.

The implications in steps 2.1, 1.2, 1.3, 1.4 and 2.2 establish all five equivalences.