Prove that $\bigcup\Bigr\{A\setminus B\;\Bigr|\,A\in\mathcal F\Bigr\}\subseteq\bigcup\Bigr(\mathcal F\setminus\mathscr P(B)\Bigr)$.

Question

Not a duplicate of

Suppose $B$ is a set and $F$ is a family of sets. Prove that $\bigcup\{A \setminus B | A \in F\} \subseteq \bigcup(F \setminus P(B))$.

This is exercise $3.4.14$ from the book How to Prove it by Velleman $($$2^{nd}$ edition$)$:

Suppose $B$ is a set and $\mathcal F$ is a family of sets. Prove that $\bigcup\Bigr\{A\setminus B\:\Bigr|\,A\in\mathcal F\Bigr\}\subseteq\bigcup\Bigr(\mathcal F\setminus\mathscr P(B)\Bigr)$.

Here is my proof:

Let $x$ be an arbitrary element of $\bigcup\Bigr\{A\setminus B\;\Bigr|\,A\in\mathcal F\Bigr\}$. So we can choose some $A_0$ such that $A_0\in \mathcal F$ and $x\in A_0\setminus B$. This means $x\in A_0$ and $x\notin B$. Since $x\in A_0$ but $x\notin B$, $\require{cancel}A_0\cancel{\subseteq} B$. Ergo $A_0\notin \mathscr P(B)$. From $A_0\in \mathcal F$ and $A_0\notin \mathscr P(B)$, $A_0\in\mathcal F\setminus\mathscr P(B)$. From $x\in A_0$ and $A_0\in\mathcal F\setminus \mathscr P(B)$, $x\in\bigcup\Bigr(\mathcal F\setminus \mathscr P(B)\Bigr)$. Therefore if $x\in \bigcup\Bigr\{A\setminus B\:\Bigr|\,A\in\mathcal F\Bigr\}$ then $x\in\bigcup\Bigr(\mathcal F\setminus \mathscr P(B)\Bigr)$. Since $x$ is arbitrary, $\forall x\Biggr(x\in\bigcup\Bigr\{A\setminus B\;\Bigr|\,A\in\mathcal F\Bigr\}\rightarrow x\in\bigcup\Bigr(\mathcal F\setminus \mathscr P(B)\Bigr)\Biggr)$ and so $\bigcup\Bigr\{A\setminus B\;\Bigr|\,A\in\mathcal F\Bigr\}\subseteq\bigcup\Bigr(\mathcal F\setminus\mathscr P(B)\Bigr)$. $Q.E.D.$

Is my proof valid$?$

Thanks for your attention.