First-order sentence involving only a symmetric binary relation with only infinite models

Let us define the formulas: $$\kappa_3(x)=\exists u\exists v(Rxu\land Rxv\land Ruv)$$ $$\kappa_4(x)=\exists u\exists v\exists w(Rxu\land Rxv\land Rxw\land Ruv\land Ruw\land Rvw)$$ $$\alpha(x)=\neg\kappa_3(x)$$ $$\beta(x)=\kappa_3(x)\land\neg\kappa_4(x)$$ $$\gamma(x)=\kappa_4(x)$$ $$\sigma(x,y)=\alpha(x)\land\alpha(y)\land\exists u\exists v(Rxu\land Ruv\land Rvy\land\beta(u)\land\gamma(v))$$ Let $\psi$ be the conjunction of the sentences: $$\forall x\forall y\forall z(\sigma(x,y)\land\sigma(y,z)\to\sigma(x,z))$$ $$\forall x\neg\sigma(x,x)$$ $$\exists x\alpha(x)$$ $$\forall x\exists y(\alpha(x)\to\sigma(x,y))$$ Plainly, $\psi$ has no finite model. On the other hand, it is a straightforward exercise to construct an infinite model of $\phi\land\psi.$

Intuition behind this example. The (irreflexive) models of $\phi$ are just (undirected) graphs. The problem is to construct an asymmetric relation on an undirected graph. Given a vertex $x$ let $f(x)$ denote the maximum number of vertices in a clique containing $x$. For vertices $x$ and $y$, define $x\lt y$ to mean that $f(x),f(y)\le2$ and there is a path $x,u,v,y$ with $f(u)=3$ and $f(v)\ge4$. Then we can write a first order sentence in the language of graph theory which says that the relation $\lt$ restricted to the set $\{x:f(x)\le2\}$ is an irreflexive transitive relation with no greatest element.