 Solution

Let $\mathcal{F}$ be the class of all finite sets. It is proper. If $\mathcal{F}$ were a set, then for every set $x$ the singleton $\{x\}$ would belong to $\mathcal{F}$, soThus $\bigcup \mathcal{F}$ would be a universal set. Separation applied to the property $z\notin z$ would then give the usual Russell contradiction.

The class $\mathcal{F}$ is not transitive. For example, $\{\omega \}$ is finite, but its member $\omega$ is infinite. It is nevertheless $\phi$-closed for every formula $\phi$: if $x$ is finite, then every subsetis finite.