[proofplan] We prove that $\beta$ satisfies the two defining properties of an ordinal. First, transitivity of $\alpha$ implies that $\beta \subset \alpha$, so every element of $\beta$ lies in the ambient well-ordered set $\alpha$. Then the transitivity of the strict order $\in$ on $\alpha$ shows that $\beta$ itself is transitive. Finally, every nonempty subset of $\beta$ is also a nonempty subset of $\alpha$, so it inherits an $\in$-least element from the well-ordering of $\alpha$. [/proofplan] [step:Use transitivity of $\alpha$ to place $\beta$ and its elements inside $\alpha$] Since $\alpha$ is an ordinal, $\alpha$ is transitive. Because $\beta \in \alpha$, transitivity of $\alpha$ gives \begin{align*} \beta \subset \alpha. \end{align*} Thus every element of $\beta$ is an element of $\alpha$. In particular, if $\delta \in \beta$, then $\delta \in \alpha$. [/step] [step:Prove that $\beta$ is transitive] Let $\delta \in \beta$ and let $\gamma \in \delta$. We must prove that $\gamma \in \beta$. From the previous step, $\delta \in \alpha$. Since $\alpha$ is transitive and $\gamma \in \delta \in \alpha$, we also have $\gamma \in \alpha$. Therefore $\gamma,\delta,\beta$ are all elements of $\alpha$. The relation $\in$ well-orders $\alpha$, hence $\in$ is a transitive strict order on $\alpha$. Applying transitivity of this order to \begin{align*} \gamma \in \delta \quad\text{and}\quad \delta \in \beta, \end{align*} we obtain $\gamma \in \beta$. Since $\delta \in \beta$ and $\gamma \in \delta$ were arbitrary, $\beta$ is transitive. [/step] [step:Restrict the well-ordering of $\alpha$ to $\beta$] Let $S$ be a nonempty subset of $\beta$. Since $\beta \subset \alpha$, we have $S \subset \alpha$. Because $\in$ well-orders $\alpha$, the set $S$ has an $\in$-least element; denote it by $s \in S$. For every $t \in S$, the leastness of $s$ in $S$ with respect to the order $\in$ on $\alpha$ gives either $s=t$ or $s \in t$. Since $S \subset \beta$, this is exactly the statement that $s$ is an $\in$-least element of $S$ when $S$ is regarded as a subset of $\beta$. Therefore every nonempty subset of $\beta$ has an $\in$-least element, and $\in$ well-orders $\beta$. [/step] [step:Conclude that $\beta$ is an ordinal] We have shown that $\beta$ is transitive and that the membership relation $\in$ well-orders $\beta$. These are precisely the defining properties of an ordinal. Hence $\beta$ is an ordinal. [/step]