A transitive proper class $\mathbf C$ is called almost universal provided that for every set $x\subseteq \mathbf C$ there exists $y\in\mathbf C$ such that $x\subseteq y$. It's a theorem that a transitive proper class which satisfies Comprehension is a model for $\mathsf {ZF}$ if and only if it's almost universal.
Here's a proof that, if a proper class satisfies $\mathsf{ZF}$, then it's almost universal. Suppose $\mathbf C\vDash \mathsf{ZF}$. Let $x\subseteq\mathbf C$ be a set. $\mathbf C$ satisfies the axiom of regularity, so it satisfies the statement that every $z\in x$ is an element of some $V_\alpha$; that is, every element of $x$ has a rank according to $\mathbf C$. Thus, let $$\alpha = \sup\{\mathrm{rank}^{\mathbf C}(z)+1\,\vert\, z\in x\}$$ Then $\mathbf C\vDash x\subseteq V_\alpha$, which is to say that $x\subseteq V_\alpha^{\mathbf C}\in\mathbf C$, so we're done.
The issue here is that I don't see where I used the assumption that $\mathbf C$ is a proper class. Sets can't be almost universal (but they can be models of $\mathsf {ZF}$ assumign the existence of e.g. a strongly inaccessible cardinal), so the above theorem shouldn't apply to sets. Where have I gone wrong? Thanks.
What does "${\bf C}\models x\subseteq V_\alpha$" mean? If $\bf C$ is just some arbitrary set or class, there's no reason to believe $\alpha\in \bf C$, in which case $\bf C$ can't even express "$x\subseteq V_\alpha$."
You're implicitly using the fact that $\bf C$ contains all the ordinals. This follows since $\bf C$ is a transitive proper class model of ZF. Specifically, we have:
Since $\bf C$ is transitive, every $\bf C$-ordinal is an ordinal.
Then since $\bf C$ is a proper class, $Ord^{\bf C}=Ord$: otherwise - since $\bf C$ is a model of ZF, and so (every element has a rank)$^{\bf C}$ - ${\bf C}$ would be contained in $V_{\sup(Ord\cap{\bf C})}$ and hence be a set.