A given ultrafilter on a compact Hausdorff space has a unique limit

The closed members of $\mathcal{U}$ have the finite-intersection property: a finite intersection remains in the filter and cannot be empty.

By compactness and [L1], choose a point $x$ in the intersection of all closed members of $\mathcal{U}$. If $X=\mathrm{\varnothing }$, no ultrafilter exists, so the universal assertion is vacuous.

For every $A\in \mathcal{U}$, its closure also belongs to $\mathcal{U}$ and contains $x$; hence every neighbourhood of $x$ meets every member of $\mathcal{U}$. Thus $x$ is a cluster point and therefore a limit by [L2].

If $x$ and $y$ were distinct limits, [L3] would give disjoint open neighbourhoods $V$ of $x$ and $W$ of $y$. Both would belong to $\mathcal{U}$, forcing $V\cap W=\mathrm{\varnothing }$ into the filter, a contradiction. Hence the limit is unique, including in a singleton space.