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 $\{\bar{A}:A\in \mathcal{F}\}$ 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 $p\in {\bigcap }_{A\in \mathcal{F}}\bar{A}$.

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 $p\in \bar{A}$; 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 $p\in \bar{C}=C$. Thus $\bigcap \mathcal{C}\mathrm{\ne }\mathrm{\varnothing }$, and [L1] gives compactness.

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