I need to prove two properties of the cumulative hiearachy
$\forall x\forall \alpha \in \mathbb{ON}\ (x \in V_{\alpha} \iff \exists \beta \in \alpha. x \subseteq V_{\beta})$
$\forall x\forall \alpha \in \mathbb{ON} \ (x \in (V_{\alpha + 1} - V_{\alpha}) \iff rank(x) = \alpha)$
Where the definition of the rank is
$rank(x)$ = sup{$rank(u)+1 : u \in x$}
For 1, I can prove the case where $\alpha$ is a limit ordinal as well as the case where $\alpha = \beta +1$ because it is obvious that $x \in V_{\alpha} \to x \subseteq V_{\beta}$ but I find it hard to generalize by induction.
For 2, I can prove that {$x : rank(x) < \alpha$} $\subseteq V_{\alpha}$ and thus since $x \notin P(\alpha)$, we have $rank(x) \ge \alpha$. But that is not enough.
I think that you just have a difficulty with understanding how to think about these inductive proofs. So let's do the first one.
The key point is that the induction hypothesis lets you have "memories of the past". If $x\in V_\alpha$, then either there is some $\gamma<\alpha$ such that $x\in V_\gamma$, or $\alpha=\gamma+1$ and then $x\subseteq V_\gamma$.
In the first case, the induction hypothesis tells you there is some $\beta<\gamma<\alpha$, such that $x\subseteq V_\beta$. In the second case, well, $\gamma=\beta$ does the work.
The point is that the inductive argument lets you "rest easy" when you can reduce the problem to a previous case. While it seems that induction is stitching an argument out parts—and that's not entirely incorrect—the main idea with induction is that you carry your past with you, and you use it to solve problems that you encountered before. Case in point, if $x\in V_\gamma$, then we already solved that issue.