Suppose that we have a set $x$ and an ordinal $\beta$ and we know that $x \in V_{\beta}$. Give a proof of the following: 1. $\{x\} \in V_{\beta +1} $ 2. $x\cup \{x\} \in V_{\beta+1}$ 3. $s(s(s(x))) \in V_{\beta+3}$ where $s(x)= x \cup \{x\}.$
I believe my proof for the first one is correct: Proof (1.) If $x\in V_{\beta}$ then we know $x\in V_{\beta+1}$ which implies $x\in \mathcal{P} (V_{\beta})$ and moreover, $x \subseteq V_{\beta}$. Thus, $\{x\} \in V_{\beta+1}$.
I am unsure on how to proceed for the others.
Your proof for 1 is a little shaky. I think every fact you state is correct but I don't quite see the reasoning between them at a couple points. Anyhow, the argument is simpler: $$ x\in V_{\beta} \implies \{x\}\subseteq V_{\beta} \implies \{x\}\in \mathcal P(V_{\beta}) = V_{\beta+1}.$$
For the second, you can show this by showing $x\cup \{x\} \subseteq V_\beta$ which means you need to show every element of $x\cup \{x\}$ is in $V_{\beta}.$ The elements of $x\cup \{x\}$ are $x$ and all of the elements of $x.$ You have already shown for part 1 that $x\in V_\beta$ so you just need to show that every element of $x$ is in $V_\beta.$ This follows from the transitivity of $V_\beta.$ If that fact is not at your disposal, you can prove it by transfinite induction. (It may help to prove and understand first that the power set of a transitive set is transitive and the union of transitive sets is transitive.)
The third follows pretty readily from the second part (you didn't use anything particulars about $\beta$ or $x$ so it's just applying part 2 three times.)