Problem. If $(x_i)$ is a net converging to a point $x$, show that $(x_i)$ converges to $x$ along any nonprincipal ultrafilter.
I will define these things below.
A directed set is a poset $(I,\leq)$ such that any two elements have an upper bound.
A net in a topological space $X$ is a function $I\rightarrow X$, where $I$ is a directed set. This is typically denoted $(x_i)_{i\in I}$ to indicate $i\mapsto x_i$. I will use $(x_i)$, omitting the index when convenient.
A (classical) limit of a net $(x_i)$ is a point $x\in X$ such that: for all neighborhoods $U$ of $x$, there exists an index $i_0\in I$ such that $x_i\in U$ for all $i\geq i_0$. Notation: $$\lim_{i\in I} x_i = x$$
Now let $p$ be an ultrafilter on $I$. An ultralimit of $(x_i)$ is a point $x\in X$ such that: for all neighborhoods $U$ of $x$, the set $\{i\in I : x_i\in U\}$ belongs to $p$. Notation: $$\lim_{i\rightarrow p} x_i = x$$ In a compact Hausdorff space, every net has a unique ultralimit along any ultrafilter, regardless of whether the classical limit exists. Now I claim that if the classical limit exists, then it is equal to the ultralimit along any (nonprincipal) ultrafilter.
Claim. If $(x_i)$ has a classical limit $x$, then $x$ is also an ultralimit of $(x_i)$ (along any nonprincipal ultrafilter).
Proof in the case of sequences
I can prove this when $(x_n)_{n\geq 1}$ is a sequence (i.e. $I=\mathbb{N}$). This simply because the "tail-end" $[i_0,\infty)$ is cofinite for all $i_0\in \mathbb{N}$, so it belongs to any nonprincipal ultrafilter. The argument generalizes to any directed set in which all tail-ends are cofinite.
Here are the details of the argument. Suppose $(x_n)$ converges classically to $x$; we must show that $(x_n)$ converges to $x$ along any nonprincipal ultrafilter $p$. For any neighborhood $U$ of $x$, there exists $n_0$ so that $x_n\in U$ for all $n\geq n_0$. Thus $\{n\geq 1 : x_n\in U\}$ contains the tail-end $[n_0,\infty)$. But since $[n_0,\infty)$ is cofinite, it belongs to $p$ since $p$ is nonprincipal. By upper-closure of ultrafilters, this implies $\{n\geq 1 : x_n\in U\}$ also belongs to $p$. QED.
I don't know how to generalize the above argument unless $(I,\leq)$ has the property that every tail-end is cofinite. This seems like a pivotal part of the argument though, so I suspect that my claim is not true in general. Any thoughts?
This is true only in the case when every tail-end of $(I,\leq)$ is cofinite, where (as you mention) the argument generalizes easily. Indeed, suppose there is some $j\in I$ such that $A=\{i\in I:i\geq j\}$ is not cofinite, so $B=\{i\in I:i\not\geq j\}$ is infinite. Consider a net $(x_i)$ which takes constant value $x$ on $A$ and constant value $y$ on $B$. Then $(x_i)$ converges to $x$, but with respect to any ultrafilter containing $B$ it converges to $y$ (and thus not does not converge to $x$ if $y\neq x$ and $X$ is Hausdorff, for instance). In particular, since $B$ is infinite, there exist nonprincipal ultrafilters which contain it.
The correct generalization is that the result applies to ultrafilters which contain every tail-end of $I$, rather than to nonprincipal ultrafilters.