 Solution

Use transfinite induction. The inclusion at zero is immediate. If $V_{\beta }\subseteq V_{\beta +1}$, monotonicity of the power set givesIf $\alpha$ is a limit and $x\in V_{\alpha }$, choose $\beta <\alpha$ with $x\in V_{\beta }$. The inductive inclusion $V_{\beta }\subseteq \mathcal{P}(V_{\beta })$ means that $x\subseteq V_{\beta }\subseteq V_{\alpha }$. Thus $x\in \mathcal{P}(V_{\alpha })=V_{\alpha +1}$. This proves $V_{\alpha }\subseteq V_{\alpha +1}$ for every ordinal. These successive inclusions and the union definition also give $V_{\beta }\subseteq V_{\alpha }$ whenever $\beta \le \alpha$.