= 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 $\mathcal F$, so
$$
x\in\bigcup\mathcal F.
$$
Thus $\bigcup\mathcal F$ 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 $\varphi$-closed for every formula $\varphi$: if $x$ is finite, then every subset
$$
\{z\in x:\varphi(z)\}
$$
is finite.
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