I do not have an instructive example of a non-parafinite congruence, so this is not an answer per se but rather an extended comment. Let me suggest two things. Firstly, it is useful to think of this in terms of so-called Leibniz congruences. Secondly, already in the case of semilattices the problem may be quite non-trivial.

Let us say that a congruence $\theta$ on an algebra $\mathbf{A}$ is a *Leibniz congruence* if there is a set $F \subseteq A$ such that $\theta$ is the largest congruence below the two-element equivalence relation $E_{F}$ naturally associated with $F$. This congruence is studied extensively in algebraic logic, where it is denoted by $\Omega^{\mathbf{A}} F$. As Benjamin Steinberg points out in the comments, in the context of semigroups these are known as syntactic congruences.

**Observation.** Each congruence $\theta$ is an intersection of Leibniz congruences: $\theta = \bigcap_{a \in A} \Omega^{\mathbf{A}} (a / \theta)$.

**Observation.** Parafinite congruences are precisely the congruences of the form $\Omega^{\mathbf{A}} F_{1} \cap \dots \cap \Omega^{\mathbf{A}} F_{n}$ for some finite family of sets $F_{1}, \dots, F_{n} \subseteq A$.

**Proof.** Each parafinite congruence $\theta_{E}$ has this form, just take the finitely many equivalence classes of $E$. Conversely, let $F_{1}, \dots, F_{n} \subseteq A$ be a finite family of sets. Let $E = E_{F_{1}} \cap \dots \cap E_{F_{n}}$ and let $G_{1}, \dots, G_{m}$ be the equivalence classes of $E$. We have $\theta_{E} = \Omega^{\mathbf{A}} G_{1} \cap \dots \cap \Omega^{\mathbf{A}} G_{m}$, so it suffices to show that $\Omega^{\mathbf{A}} G_{1} \cap \dots \cap \Omega^{\mathbf{A}} G_{m} = \Omega^{\mathbf{A}} F_{1} \cap \dots \cap \Omega^{\mathbf{A}} F_{n}$. The left-to-right inclusion holds because each $F_{i}$ is a union of some subfamily of the family $G_{j}$ and in general $\Omega^{\mathbf{A}} X_{1} \cap \dots \cap \Omega^{\mathbf{A}} X_{k} \subseteq \Omega^{\mathbf{A}} (X_{1} \cup \dots X_{k})$. The right-to-left inclusion holds because each $G_{j}$ is a Boolean combination of the sets $F_{i}$ and in addition to the previous observation concerning unions we also have the observation $\Omega^{\mathbf{A}} X = \Omega^{\mathbf{A}} (A - X)$ concerning complements and $\Omega^{\mathbf{A}} X_{1} \cap \dots \cap \Omega^{\mathbf{A}} X_{k} \subseteq \Omega^{\mathbf{A}} (X_{1} \cap \dots \cap X_{k})$ concerning intersections.

**Corollary.** Each dually compact congruence is parafinite.

An interesting open problem posed by Josep Maria Font and my colleague Tommaso Moraschini here is the following.

**Open problem.** Is each congruence on a semilattice a Leibniz congruence?

This seems like it should be quite a simple problem, but it has not turned out that way so far. Leibniz congruences being a special case of parafinite congruences, it makes sense to ask the following variant question, which I suspect may also turn out to be quite non-trivial.

**Question.** Is each congruence on a semilattice parafinite?

In case you make any progress in this direction, I would certainly be interested in hearing about it. If I had to make a wild guess, I would guess that such a congruence will exist but might be quite complicated to construct.