I am given the problem:
If $X$ is an inductive set, then the set $\{x\in X: x\notin x \wedge x \text{ is transitive} \} = S$ is inductive.
We are given the definition
$T$ is transitive if $x \in T$ implies $x \subset T$
$\emptyset \in S$ since $\emptyset \in X$, $\emptyset \notin \emptyset$, and $\emptyset$ is transitive.
Now let $x \in S$. Clearly, $x \cup \{x\}$ must be in $X$ by assumption. So, if we take some $y \in x\cup\{x\}$, then $y = x$ or $y\in x$. If it's the first case, then use the fact that $x$ is itself transitive so $y \subset x$. If it's the later, then $y = x \subset x\cup \{x\}$. Therefore $x \cup \{x\}$ is transitive.
So we clearly have $x\cup\{x\} \neq x$ since $x \notin x$. Now assume $x\cup\{x\} =y$ for some $y\in x$. Since $x$ is transitive, $y\subset x$. But again, $x\cup\{x\}$ cannot be a subset since this inclusion $x \subset x\cup\{x\}$ is proper. Therefore, $x \cup \{x\}$ is not equal to any element in $x \cup\{x\}$ and is not an element.
Therefore, we showed that if $x \in S$, $x\cup\{x\}$ will also be in $S$. Hence, $S$ is inductive.
Is my attempt correct?