[proofplan] Choose a well-order on $A$ and pass to its order type $\alpha$, so that $A$ is equinumerous with an ordinal. Among the ordinals at most $\alpha$ that are equinumerous with $A$, take the least one and call it $\kappa$. This ordinal is exactly $|A|$ by definition, and its minimality rules out any smaller ordinal equinumerous with it. [/proofplan] [step:Choose an ordinal representative of the well-order type of $A$] Since $A$ is well-orderable, there exists a well-order relation $<_{A}$ on $A$. Let $\alpha$ be the order type of the well-ordered set $(A, <_{A})$. By the definition of order type, there exists an order isomorphism \begin{align*} f: (A, <_{A}) &\to (\alpha, \in). \end{align*} In particular, $f: A \to \alpha$ is a bijection of sets, so $A \approx \alpha$. [/step] [step:Select the least ordinal equinumerous with $A$] Define \begin{align*} S := \{\beta \in \alpha + 1 : \beta \approx A\}. \end{align*} The set $S$ is nonempty because $\alpha \in S$, as $A \approx \alpha$. Since $S$ is a nonempty set of ordinals, it has a least element under the ordinal order. Denote this least element by $\kappa$. We claim that $\kappa$ is the least ordinal equinumerous with $A$. Indeed, $\kappa \in S$, so $\kappa \approx A$. If $\gamma$ is any ordinal with $\gamma \approx A$ and $\gamma < \kappa$, then $\gamma < \kappa \leq \alpha$, hence $\gamma \in \alpha + 1$. Therefore $\gamma \in S$, contradicting the minimality of $\kappa$ in $S$. Thus no ordinal smaller than $\kappa$ is equinumerous with $A$. By the definition of cardinality for well-orderable sets, the least ordinal equinumerous with $A$ is $|A|$. Hence $\kappa = |A|$, and since $\kappa \approx A$, we obtain $A \approx |A|$. [/step] [step:Use minimality to prove that $|A|$ is initial] An ordinal $\lambda$ is initial precisely when no ordinal $\beta < \lambda$ is equinumerous with $\lambda$. We prove this for $\lambda = |A| = \kappa$. Suppose, for contradiction, that there exists an ordinal $\beta < \kappa$ such that $\beta \approx \kappa$. Since $\kappa \approx A$, transitivity of equinumerosity gives $\beta \approx A$. This contradicts the fact proved above that no ordinal smaller than $\kappa$ is equinumerous with $A$. Therefore no ordinal $\beta < \kappa$ satisfies $\beta \approx \kappa$, so $\kappa$ is an initial ordinal. Since $\kappa = |A|$, the cardinal $|A|$ is an initial ordinal, and the previously established bijection gives $A \approx |A|$. [/step]