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 
$$
\forall x\ (x\in T\implies P(x))\ ?
$$

This is true.
Assume to the contrary that there is an ordinal $\alpha$ such that no set equinumerous to $\alpha$ is contained in $P$. Then there is a smallest such $\alpha$. However this means that for each $\beta<\alpha$ we can choose an $A_\beta$ such that $A_\beta \subseteq P$ and $|A_\beta|=|\beta|$ (employing Scott's trick to cut the space of candidate $A_\beta$s down to set size before choosing frely). But then $\bigcup_{\beta<\alpha} A_\beta \subseteq P$ and its cardinality is at least $\alpha$.
