= Solution
{c}
We prove simultaneously that each $L_\alpha$ is a <transitive set> and that $L_\alpha\subseteq L_{\alpha+1}$. The key observation is that if $A$ is transitive and $x\in A$, then $x\subseteq A$ and
$$
x=\{y\in A:(A,\in)\models y\in x\}.
$$
This is an internal definition with parameter $x$, so $x\in\operatorname{Def}(A,1)$. Hence $A\subseteq\operatorname{Def}(A,1)$. Every $z\in\operatorname{Def}(A,1)$ is a subset of $A$; if $y\in z$, then $y\in A\subseteq\operatorname{Def}(A,1)$. Thus the definable power set is transitive too.
Start with the empty set. At successor stages the observation proves transitivity and inclusion. At a limit stage the union of the earlier transitive sets is transitive, and the same observation then proves its inclusion in its own definable power set. This completes the <transfinite induction> and gives
$$
\boxed{\forall\alpha\quad L_\alpha\text{ is transitive}.}
$$
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