@Dong:
As I pointed out in comments, and Pete mentioned in his answer, well-orderability is all you need, so yes, if $P$ is a property of cardinals as you describe, then it holds of all cardinals. You can see this by taking $P'(\alpha)$ to be: "$\alpha$ is not a cardinal, or else, it is a cardinal and $P(\alpha)$", and applying the usual transfinite induction theorem to $P'$.
In 1., I took "cardinal" to mean "initial-ordinal". Several questions come to mind immediately if we remove choice from the picture:
a. Suppose $P$ is a property of (not necessarily well-ordered) cardinalities, and it has the property you mention: If for all smaller sizes it holds, it holds for the size under consideration.
This does not suffice for $P$ to hold at all sizes. For example, suppose there is an infinite Dedekind-finite set, and let $P(X)$ be the statement:
"Either $X$ is not infinite Dedekind-finite, or else, the cardinalities are well-founded below the size of $X$".
Being Dedekind-finite means that any proper subset has smaller size. It follows that if any smaller set satisfies $P$, so does $X$. But $P(X)$ is plain false if $X$ is infinite Dedekind-finite (and it is consistent with ZF that there are such sets).
b. Suppose that all sets of smaller cardinality than that of $X$ are well-orderable. It does not follow that $X$ itself is well-orderable. For example, consider Solovay's model, and take $X={\mathbb R}$. Since the perfect set property holds, any subset of the reals either has the same size as ${\mathbb R}$, or else it is countable, in which case it is well-orderable. But ${\mathbb R}$ itself is not well-orderable.
c. A significantly harder question is whether, if cardinalities are well-founded, then choice holds. This is an open problem. It was asked independently by T. Forster and D. Savaliev, and I have thought about it on and off for a while.