Does a proper class have arbitrary large subsets? Assume that we are working in ZFC, that we have a well-formed formula $P(x)$, that $x$ is the only free variable of $P(x)$, that there is no set $S$ such that 
$$
\forall x\ (x\in S\iff P(x)),
$$ 
and that $\alpha$ is a cardinal.
Is there necessarily a set $T$ of cardinality $\alpha$ such that 
$$
\forall x\ (x\in T\implies P(x))\ ?
$$

In Godel-Bernays + Proper class choice, I would say yes.
In fact, consider your proper class $A$. Since it is not empty, there is an element inside, call it $x_0$. Now consider the proper class of elements that are not $x_0$, call it $x_0^{\bot}$ and intersect it with $A$. This cannot be $A$, so there is another element in $A$ which is not $x_0$, call it $x_1$. By induction I get an injection of ordinals into $A$.
I do believe that this proof can be reformulated in ZFC + Universal choice, but I do not feel comfortable in doing it.
