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 $$ \forall x\ (x\in T\implies P(x))\ ? $$
On
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.
On
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 $\alpha$.
The answer is yes.
Let $P_\alpha=\{x\in V_\alpha\mid P(x)\}$. Namely, we intersect the class given by $P$ with each $V_\alpha$. So we get that $P(x)\iff\exists\alpha. x\in P_\alpha$.
Define by recursion a function $F(0)=0$, and $F(\alpha+1)=\beta$ if and only if $P_\beta$ is the first time where $P_{F(\alpha)}\subsetneq P_\beta$. Namely, the first time we add new members. For limit steps, take limits.
Now it follows that if $\kappa$ is any cardinal, then $P_{F(\kappa)}$ must have at least $\kappa$ elements. For any $\kappa$. So given any cardinal $\lambda$, we can carve out some subset of $P_{F(\lambda)}$ with the right cardinality.