[proofplan] We first prove, by transfinite induction, the auxiliary facts that every level $L_\alpha$ is transitive and that $L_\alpha \subset L_{\alpha+1}$. The successor step uses the defining property of $\operatorname{Def}(M)$: its elements are definable subsets of $M$, and parameters from $M$ may be used. Once adjacent inclusions are known at every stage, the inclusion $L_\alpha \subset L_\beta$ follows by a second transfinite induction on $\beta$, with limit stages handled by the union definition of $L_\beta$. [/proofplan]

[step:Prove transitivity and adjacent inclusion simultaneously]We prove by transfinite induction on the ordinal $\alpha$ the following two assertions: \begin{align*} \mathrm{T}(\alpha) &: \quad L_\alpha \text{ is transitive},\\ \mathrm{S}(\alpha) &: \quad L_\alpha \subset L_{\alpha+1}. \end{align*} Here transitivity means that for every $x \in L_\alpha$ and every $y \in x$, one has $y \in L_\alpha$. For $\alpha = 0$, we have $L_0 = \varnothing$. Thus $L_0$ is transitive, and $L_0 \subset L_1$. Assume now that $\alpha = \gamma + 1$ and that $\mathrm{T}(\gamma)$ and $\mathrm{S}(\gamma)$ hold. We first prove $\mathrm{T}(\gamma+1)$. Let $x \in L_{\gamma+1}$. By definition, \begin{align*} L_{\gamma+1} = \operatorname{Def}(L_\gamma), \end{align*} so $x$ is a definable subset of $L_\gamma$. Hence $x \subset L_\gamma$. Since $\mathrm{S}(\gamma)$ gives $L_\gamma \subset L_{\gamma+1}$, we obtain $x \subset L_{\gamma+1}$. Therefore $L_{\gamma+1}$ is transitive. We next prove $\mathrm{S}(\gamma+1)$. Let $x \in L_{\gamma+1}$. Since $L_{\gamma+1}$ is transitive, $x \subset L_{\gamma+1}$. Define the formula $\varphi(u,p)$ in the language of set theory by \begin{align*} \varphi(u,p) \quad \text{means} \quad u \in p. \end{align*} Using the parameter $x \in L_{\gamma+1}$, the subset of $L_{\gamma+1}$ defined by $\varphi(u,x)$ is exactly $x$, because $x \subset L_{\gamma+1}$. Hence $x \in \operatorname{Def}(L_{\gamma+1}) = L_{\gamma+2}$. Thus $L_{\gamma+1} \subset L_{\gamma+2}$. Finally suppose $\lambda$ is a limit ordinal and that $\mathrm{T}(\gamma)$ and $\mathrm{S}(\gamma)$ hold for every $\gamma < \lambda$. We first prove $\mathrm{T}(\lambda)$. Let $x \in L_\lambda$. By definition of the limit stage, there is some $\gamma < \lambda$ such that $x \in L_\gamma$. Since $L_\gamma$ is transitive, $x \subset L_\gamma$. Since $L_\gamma \subset L_\lambda$ by the definition \begin{align*} L_\lambda = \bigcup_{\delta < \lambda} L_\delta, \end{align*} we get $x \subset L_\lambda$. Therefore $L_\lambda$ is transitive. We now prove $\mathrm{S}(\lambda)$. Let $x \in L_\lambda$. Since $L_\lambda$ is transitive, $x \subset L_\lambda$. Again use the formula $\varphi(u,p)$ given by $u \in p$. With parameter $x \in L_\lambda$, this formula defines precisely the subset $x$ of $L_\lambda$. Therefore \begin{align*} x \in \operatorname{Def}(L_\lambda) = L_{\lambda+1}. \end{align*} Thus $L_\lambda \subset L_{\lambda+1}$. By transfinite induction, $\mathrm{T}(\alpha)$ and $\mathrm{S}(\alpha)$ hold for every ordinal $\alpha$.[/step]

[guided]We prove two facts at once because each one feeds the other. To show $L_\alpha \subset L_{\alpha+1}$, we want to say that an element $x \in L_\alpha$ is definable over $L_\alpha$ using itself as a parameter by the formula $u \in x$. But this only defines $x$ as a subset of $L_\alpha$ if $x \subset L_\alpha$, which is exactly transitivity. Thus we prove transitivity of each level and adjacent inclusion simultaneously. For the base case, $L_0 = \varnothing$. The empty set is transitive because there is no $x \in \varnothing$ to check, and $\varnothing \subset L_1$. Now let $\alpha = \gamma+1$, and assume $L_\gamma$ is transitive and $L_\gamma \subset L_{\gamma+1}$. We first show that $L_{\gamma+1}$ is transitive. If $x \in L_{\gamma+1}$, then \begin{align*} L_{\gamma+1} = \operatorname{Def}(L_\gamma). \end{align*} By definition, every element of $\operatorname{Def}(L_\gamma)$ is a subset of $L_\gamma$ definable over $L_\gamma$ with parameters from $L_\gamma$. Hence $x \subset L_\gamma$. The induction hypothesis gives $L_\gamma \subset L_{\gamma+1}$, so $x \subset L_{\gamma+1}$. This proves that $L_{\gamma+1}$ is transitive. Next we prove $L_{\gamma+1} \subset L_{\gamma+2}$. Let $x \in L_{\gamma+1}$. Since we have just proved that $L_{\gamma+1}$ is transitive, every element of $x$ lies in $L_{\gamma+1}$, so $x \subset L_{\gamma+1}$. Consider the first-order formula \begin{align*} \varphi(u,p) \quad \text{means} \quad u \in p. \end{align*} When interpreted over $L_{\gamma+1}$ with parameter $p=x$, this formula defines exactly the subset $x$ of $L_{\gamma+1}$. The parameter is allowed because $x \in L_{\gamma+1}$. Therefore $x$ belongs to $\operatorname{Def}(L_{\gamma+1})$, which is $L_{\gamma+2}$. Since $x$ was arbitrary, $L_{\gamma+1} \subset L_{\gamma+2}$. Now let $\lambda$ be a limit ordinal, and assume the two assertions hold below $\lambda$. If $x \in L_\lambda$, then by the limit definition \begin{align*} L_\lambda = \bigcup_{\gamma < \lambda} L_\gamma, \end{align*} there exists $\gamma < \lambda$ with $x \in L_\gamma$. By the induction hypothesis, $L_\gamma$ is transitive, so $x \subset L_\gamma$. Since $L_\gamma \subset L_\lambda$ by the same union definition, we have $x \subset L_\lambda$. Hence $L_\lambda$ is transitive. Finally, if $x \in L_\lambda$, then transitivity of $L_\lambda$ gives $x \subset L_\lambda$. The formula $\varphi(u,p)$ meaning $u \in p$, with parameter $x \in L_\lambda$, defines $x$ over $L_\lambda$. Therefore \begin{align*} x \in \operatorname{Def}(L_\lambda) = L_{\lambda+1}. \end{align*} Thus $L_\lambda \subset L_{\lambda+1}$. This completes the simultaneous transfinite induction.[/guided]

[step:Iterate adjacent inclusion to compare arbitrary stages]Fix an ordinal $\alpha$. We prove by transfinite induction on ordinals $\beta \geq \alpha$ that $L_\alpha \subset L_\beta$. If $\beta = \alpha$, then $L_\alpha \subset L_\alpha$ by reflexivity of inclusion. Suppose $\beta = \delta + 1$ and $\alpha \leq \delta$. By the induction hypothesis, $L_\alpha \subset L_\delta$. From the previous step, $L_\delta \subset L_{\delta+1} = L_\beta$. By transitivity of subset inclusion, \begin{align*} L_\alpha \subset L_\beta. \end{align*} Finally suppose $\beta$ is a limit ordinal and $\alpha < \beta$. Since \begin{align*} L_\beta = \bigcup_{\delta < \beta} L_\delta, \end{align*} and since $\alpha < \beta$, we have $L_\alpha \subset L_\beta$ directly from the definition of union. Therefore, for all ordinals $\alpha$ and $\beta$, if $\alpha \leq \beta$, then $L_\alpha \subset L_\beta$.[/step]

[guided]Once we know every adjacent inclusion $L_\delta \subset L_{\delta+1}$, the general monotonicity statement is obtained by iterating these inclusions through the ordinal interval from $\alpha$ to $\beta$. Fix $\alpha$. We prove that $L_\alpha \subset L_\beta$ for every $\beta \geq \alpha$ by transfinite induction on $\beta$. If $\beta=\alpha$, there is nothing to build: every set is a subset of itself, so $L_\alpha \subset L_\alpha$. Now suppose $\beta=\delta+1$ and $\alpha \leq \delta$. The induction hypothesis gives \begin{align*} L_\alpha \subset L_\delta. \end{align*} The first step of the proof gives the adjacent inclusion \begin{align*} L_\delta \subset L_{\delta+1}. \end{align*} Since $\beta=\delta+1$, transitivity of subset inclusion yields \begin{align*} L_\alpha \subset L_\delta \subset L_{\delta+1} = L_\beta. \end{align*} This proves the successor case. It remains to handle the case where $\beta$ is a limit ordinal. If $\alpha < \beta$, then $\alpha$ is one of the indices appearing in the union defining $L_\beta$: \begin{align*} L_\beta = \bigcup_{\delta < \beta} L_\delta. \end{align*} Therefore every element of $L_\alpha$ belongs to this union, and hence $L_\alpha \subset L_\beta$. Combining the base, successor, and limit cases proves the desired monotonicity for all ordinals $\alpha \leq \beta$.[/guided]