[proofplan] We first prove, by transfinite induction on the stage $\alpha$, the structural facts that each $V_\alpha$ is transitive, that $V_\alpha \subset V_{\alpha+1}$, and that every element of $V_\alpha$ is already a subset of some earlier stage. The successor case uses the identity $V_{\delta+1}=\mathcal{P}(V_\delta)$, while the limit case uses the union defining $V_\lambda$. Once the one-step inclusions $V_\alpha \subset V_{\alpha+1}$ are available, monotonicity for arbitrary $\alpha \leq \beta$ follows by a second transfinite induction on $\beta$. [/proofplan]

[step:Prove transitivity, one-step inclusion, and rank bounds simultaneously]We prove by transfinite induction on an ordinal $\alpha$ the following three assertions: \begin{align*} \mathrm{T}(\alpha)&:\quad \text{if }x \in y\in V_\alpha,\text{ then }x\in V_\alpha,\\ \mathrm{I}(\alpha)&:\quad V_\alpha \subset V_{\alpha+1},\\ \mathrm{R}(\alpha)&:\quad \text{if }x\in V_\alpha,\text{ then there exists }\gamma\in\alpha\text{ such that }x\subset V_\gamma. \end{align*} For $\alpha=0$, the assertions $\mathrm{T}(0)$ and $\mathrm{R}(0)$ are vacuous because $V_0=\varnothing$. Also $V_0=\varnothing\subset \mathcal{P}(\varnothing)=V_1$, so $\mathrm{I}(0)$ holds. Now suppose $\alpha=\delta+1$ and that $\mathrm{T}(\delta)$, $\mathrm{I}(\delta)$, and $\mathrm{R}(\delta)$ hold. If $y\in V_{\delta+1}$, then $y\in \mathcal{P}(V_\delta)$, so $y\subset V_\delta$. Hence, whenever $x\in y$, we have $x\in V_\delta$. By $\mathrm{I}(\delta)$, $V_\delta\subset V_{\delta+1}$, and therefore $x\in V_{\delta+1}$. This proves $\mathrm{T}(\delta+1)$. If $x\in V_{\delta+1}$, then $x\subset V_\delta$. Since $\delta\in\delta+1$, this proves $\mathrm{R}(\delta+1)$. To prove $\mathrm{I}(\delta+1)$, let $x\in V_{\delta+1}$. The preceding sentence gives $x\subset V_\delta$. By $\mathrm{I}(\delta)$, $V_\delta\subset V_{\delta+1}$, hence $x\subset V_{\delta+1}$. Therefore $x\in \mathcal{P}(V_{\delta+1})=V_{\delta+2}$. Finally suppose $\lambda$ is a limit ordinal and that $\mathrm{T}(\delta)$, $\mathrm{I}(\delta)$, and $\mathrm{R}(\delta)$ hold for every $\delta\in\lambda$. Let $y\in V_\lambda$. By definition of $V_\lambda$, there exists $\delta\in\lambda$ such that $y\in V_\delta$. If $x\in y$, then $\mathrm{T}(\delta)$ gives $x\in V_\delta$. Since $\delta\in\lambda$, we have $V_\delta\subset \bigcup_{\eta\in\lambda}V_\eta=V_\lambda$ by the definition of union. Thus $x\in V_\lambda$, proving $\mathrm{T}(\lambda)$. To prove $\mathrm{I}(\lambda)$, let $x\in V_\lambda$. Then $x\in V_\delta$ for some $\delta\in\lambda$. If $z\in x$, then $\mathrm{T}(\delta)$ gives $z\in V_\delta$, and hence $z\in V_\lambda$. Thus $x\subset V_\lambda$, so $x\in \mathcal{P}(V_\lambda)=V_{\lambda+1}$. To prove $\mathrm{R}(\lambda)$, let $x\in V_\lambda$. Choose $\delta\in\lambda$ such that $x\in V_\delta$. Since $V_0=\varnothing$, this $\delta$ is not $0$. Applying $\mathrm{R}(\delta)$, there exists $\gamma\in\delta$ such that $x\subset V_\gamma$. Because $\gamma\in\delta\in\lambda$ and $\lambda$ is transitive as an ordinal, $\gamma\in\lambda$. Therefore $x\subset V_\gamma$ for some $\gamma\in\lambda$. By transfinite induction, $\mathrm{T}(\alpha)$, $\mathrm{I}(\alpha)$, and $\mathrm{R}(\alpha)$ hold for every ordinal $\alpha$.[/step]

[guided]We prove three properties together because the arguments depend on each other. The three properties are: \begin{align*} \mathrm{T}(\alpha)&:\quad \text{if }x \in y\in V_\alpha,\text{ then }x\in V_\alpha,\\ \mathrm{I}(\alpha)&:\quad V_\alpha \subset V_{\alpha+1},\\ \mathrm{R}(\alpha)&:\quad \text{if }x\in V_\alpha,\text{ then there exists }\gamma\in\alpha\text{ such that }x\subset V_\gamma. \end{align*} The base stage is $\alpha=0$. Since $V_0=\varnothing$, there are no elements $y\in V_0$ and no elements $x\in V_0$. Therefore the transitivity assertion $\mathrm{T}(0)$ and the rank-bound assertion $\mathrm{R}(0)$ have no cases to check. For the one-step inclusion, $V_1=\mathcal{P}(V_0)=\mathcal{P}(\varnothing)$, and $\varnothing\subset\mathcal{P}(\varnothing)$, so $V_0\subset V_1$. Now consider a successor stage $\alpha=\delta+1$. The defining identity is \begin{align*} V_{\delta+1}=\mathcal{P}(V_\delta). \end{align*} Thus an element of $V_{\delta+1}$ is exactly a subset of $V_\delta$. Let $y\in V_{\delta+1}$ and let $x\in y$. Since $y\in\mathcal{P}(V_\delta)$, we have $y\subset V_\delta$, so $x\in V_\delta$. The induction hypothesis $\mathrm{I}(\delta)$ says $V_\delta\subset V_{\delta+1}$, hence $x\in V_{\delta+1}$. This proves that $V_{\delta+1}$ is transitive. The rank-bound assertion at the same successor stage is built into the definition. If $x\in V_{\delta+1}$, then $x\in\mathcal{P}(V_\delta)$, so $x\subset V_\delta$. Since $\delta\in\delta+1$, the required earlier stage is $\gamma=\delta$. It remains at the successor stage to prove $V_{\delta+1}\subset V_{\delta+2}$. Let $x\in V_{\delta+1}$. From the previous paragraph, $x\subset V_\delta$. By the induction hypothesis $\mathrm{I}(\delta)$, every element of $V_\delta$ is an element of $V_{\delta+1}$. Therefore $x\subset V_{\delta+1}$. By the definition of the next stage, \begin{align*} V_{\delta+2}=\mathcal{P}(V_{\delta+1}), \end{align*} so $x\in V_{\delta+2}$. Now let $\lambda$ be a limit ordinal. The defining identity is \begin{align*} V_\lambda=\bigcup_{\delta\in\lambda}V_\delta. \end{align*} To prove transitivity, take $y\in V_\lambda$ and $x\in y$. Since $V_\lambda$ is a union of the earlier stages, there is some $\delta\in\lambda$ with $y\in V_\delta$. By the induction hypothesis $\mathrm{T}(\delta)$, the implication $x\in y\in V_\delta$ gives $x\in V_\delta$. Since $V_\delta$ is one of the sets appearing in the union defining $V_\lambda$, we get $x\in V_\lambda$. To prove $V_\lambda\subset V_{\lambda+1}$, let $x\in V_\lambda$. Choose $\delta\in\lambda$ such that $x\in V_\delta$. If $z\in x$, then transitivity of $V_\delta$ gives $z\in V_\delta$, hence $z\in V_\lambda$. Therefore every element of $x$ lies in $V_\lambda$, meaning $x\subset V_\lambda$. Since $V_{\lambda+1}=\mathcal{P}(V_\lambda)$, this gives $x\in V_{\lambda+1}$. Finally, let $x\in V_\lambda$. Choose $\delta\in\lambda$ such that $x\in V_\delta$. The case $\delta=0$ cannot occur because $V_0=\varnothing$. Therefore the induction hypothesis $\mathrm{R}(\delta)$ applies to $x\in V_\delta$ and gives an ordinal $\gamma\in\delta$ such that $x\subset V_\gamma$. Since ordinals are transitive sets, $\gamma\in\delta\in\lambda$ implies $\gamma\in\lambda$. Thus $x$ is a subset of an earlier stage $V_\gamma$ with $\gamma\in\lambda$. The base, successor, and limit cases prove the three assertions for every ordinal $\alpha$ by transfinite induction.[/guided]

[step:Deduce monotonicity from the one-step inclusions]We prove that $V_\alpha\subset V_\beta$ whenever $\alpha\leq\beta$ by transfinite induction on $\beta$. For $\beta=0$, the only ordinal $\alpha$ with $\alpha\leq 0$ is $\alpha=0$, and $V_0\subset V_0$. Suppose $\beta=\delta+1$ and the assertion holds for $\delta$. Let $\alpha\leq\delta+1$. If $\alpha=\delta+1$, then $V_\alpha=V_\beta$. If $\alpha\leq\delta$, the induction hypothesis gives $V_\alpha\subset V_\delta$, while the already proved one-step inclusion gives $V_\delta\subset V_{\delta+1}=V_\beta$. Hence $V_\alpha\subset V_\beta$. Suppose $\beta=\lambda$ is a limit ordinal and the assertion holds for every $\delta\in\lambda$. Let $\alpha\leq\lambda$. If $\alpha=\lambda$, then $V_\alpha=V_\beta$. If $\alpha\in\lambda$, then for every $x\in V_\alpha$, the set $V_\alpha$ is one of the sets in the union defining $V_\lambda$, so $x\in V_\lambda$. Thus $V_\alpha\subset V_\lambda=V_\beta$. Therefore $V_\alpha\subset V_\beta$ for all ordinals $\alpha\leq\beta$.[/step]

[guided]We now turn the local inclusion $V_\delta\subset V_{\delta+1}$ into full monotonicity. The statement to prove is: \begin{align*} \alpha\leq\beta \quad\Longrightarrow\quad V_\alpha\subset V_\beta. \end{align*} We prove this by transfinite induction on the larger ordinal $\beta$. When $\beta=0$, the only ordinal $\alpha$ satisfying $\alpha\leq0$ is $\alpha=0$. Hence $V_\alpha=V_0=V_\beta$, and the inclusion holds. Now assume $\beta=\delta+1$. Let $\alpha\leq\delta+1$. There are two cases. If $\alpha=\delta+1$, then $V_\alpha=V_\beta$, so the inclusion is equality. If $\alpha\leq\delta$, then the induction hypothesis at $\delta$ gives $V_\alpha\subset V_\delta$. From the previous step, we already know the one-step inclusion $V_\delta\subset V_{\delta+1}$. Combining these inclusions gives \begin{align*} V_\alpha\subset V_\delta\subset V_{\delta+1}=V_\beta. \end{align*} Now assume $\beta=\lambda$ is a limit ordinal. Let $\alpha\leq\lambda$. If $\alpha=\lambda$, the inclusion is equality. If $\alpha\in\lambda$, then $V_\lambda$ is defined as the union of all earlier stages: \begin{align*} V_\lambda=\bigcup_{\delta\in\lambda}V_\delta. \end{align*} Since $\alpha\in\lambda$, the set $V_\alpha$ appears as one member of this union. Therefore every $x\in V_\alpha$ is automatically an element of $V_\lambda$. This proves $V_\alpha\subset V_\lambda=V_\beta$. The induction covers the zero, successor, and limit cases, so monotonicity holds for every pair of ordinals $\alpha\leq\beta$.[/guided]

[step:Assemble the three conclusions] The first conclusion is the monotonicity just proved: if $\alpha\leq\beta$, then $V_\alpha\subset V_\beta$. The second conclusion is $\mathrm{T}(\alpha)$ for arbitrary $\alpha$: if $x\in y\in V_\alpha$, then $x\in V_\alpha$. The third conclusion is $\mathrm{R}(\alpha)$ for arbitrary $\alpha$: if $x\in V_\alpha$, then there exists $\gamma\in\alpha$ such that $x\subset V_\gamma$. These are exactly the three asserted basic properties of the cumulative hierarchy. [/step]