Suppose that $x_{i \in I}$ is a generalized sequence on a compact Hausdorff space $X$, indexed by the directed set $I$, and with the property that $\exists \, j \in I$ such that $\forall i \geqslant j$, then $x_i = k$, where $k \in X$. We then pick a non-principal ultrafilter $U$ on the index set $I$ (so $U \subset \mathcal{P}(I)$).
My claim is that we then have $\lim_{U} (x_i) = k$.
First we notice that $X$ being compact and Hausdorff, the ultralimit, $\lim_U (x_i)$, of this sequence always exists uniquely. Also, we can just assume that such an ultrafilter on $I$ exists ($I$ being an infinite index is actually sufficient and necessary for that).
I have been unable so far to prove that this claim is true and I am starting to think I am missing an assumption for it to be true.
This isn't true. For instance, let $I=\mathbb{N}\cup\{\infty\}$; then any sequence $(x_i)_{i\in I}$ is eventually constant with value $k=x_\infty$ (since you can always just take $j=\infty$). But if, say, there is a $y\in X$ distinct from $k$ such that $x_n=y$ for all $n\in \mathbb{N}$, then $\lim_U(x_i)=y$, not $k$.
For the result to be true, you could require that $U$ contains the set $\{i\in I:i\geq j\}$ for all $j\in I$. Note, however, that there may be no nonprincipal ultrafilter with this property, as the previous example (or more generally, any directed set $I$ which has a greatest element) shows.