= Solution
The <Axiom of foundation> states that every nonempty set $A$ has a member $a$ disjoint from $A$. The principle of membership induction states that a property $P$ satisfying $\forall x[(\forall y\in x\ P(y))\Rightarrow P(x)]$ holds for every set.
Assume Foundation and suppose a progressive property fails at a set $x$. In the transitive closure of $\{x\}$, form the nonempty set of counterexamples by Separation. Foundation selects a counterexample $a$ with no counterexample among its members. Transitivity makes those members belong to the closure, so each satisfies $P$, and progressiveness gives $P(a)$, a contradiction. Conversely, if a nonempty $A$ had no member disjoint from it, the property $P(x):x\notin A$ would be progressive: a member of $A$ necessarily has a member in $A$. Membership induction would then say no set belongs to $A$, a contradiction. Thus \b[Foundation and membership induction are equivalent].
Apply induction to the property that a set belongs to some stage of the <cumulative hierarchy>. For each member $y$ of $x$, choose its least stage ordinal. Replacement collects these ordinals into a set; take an ordinal $\gamma$ large enough that all members lie in $V_\gamma$. Then $x\subseteq V_\gamma$ and $x\in V_{\gamma+1}$. This proves every set appears in the hierarchy.
For finite stages let $a_n=|V_n|$. Then $\boxed{a_0=0,\quad a_{n+1}=2^{a_n}}$, giving $0,1,2,4,16,65536,\ldots$. The stage $V_\omega$ is countably infinite, as the union of the finite stages. Thus $|V_{\omega+1}|=2^{\aleph_0}=|\mathbb R|$. Cantor's theorem makes $|V_{\omega+2}|$ strictly larger, and every later stage contains it. Earlier stages have at most countable size. Therefore $\boxed{|V_\alpha|=|\mathbb R|\iff\alpha=\omega+1}$.
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