There's no set $X$ such that $P(X)\subset X$. [duplicate]

I want to show the above assertion, I guess I am not sure what's the definition of a set, I thought it's the basic building block.

Anyway, if the above titled assertion weren't true then for every $Y \subset X$ we will also have $Y \in X$, which should contradict $X$ being a set, but why is that?

If you know that no set can be an element of itself, then you can show that $P(X)$ is not a subset of $X$ by simply observing that $X\in P(X)$ while $X\notin X.$

If you don't know that, then you can still prove that $P(X)$ is not a subset of $X$ by using Bertrand Russell's trick: define the set $R=\{x\in X:x\notin x\},$ the set of all elements of $X$ which are not elements of themselves. (If no set is an element of itself, then $R=X$.) Then $R\in P(X)$ but $R\notin X,$ since assuming $R\in X$ leads to the contradiction $R\in R\leftrightarrow R\notin R.$

If $\mathcal P(X)\subset X$ then $\forall Y:(Y\subseteq X\to Y\in X)$.   That is, all subsets of $X$ would be elements of $X$.

Now consider that $X$ is an element of the powerset of $X$ (ie $X\subseteq X$).   So...