I'm self learning from the book "How to Prove it" by Velleman (3rd edition). I don't have access to a math professor, so I need a little help from the community. Please verify my proof.
Problem 17a, pag. 151
Suppose $\mathcal F$ is a nonempty family of sets and $B$ is a set. Prove that $B \cup(\bigcup \mathcal F) = \bigcup(\mathcal F \cup \{B\})$.
Proof. Suppose $x \in B \cup (\bigcup \mathcal F)$, then $x \in B$ or $x \in \bigcup \mathcal F$. So there are two cases:
Case 1. $x \in B$, because $B \in (\mathcal F \cup \{B\})$, then definitely $x \in \bigcup(\mathcal F \cup \{B\})$
Case 2. $x \in \bigcup \mathcal F$, then $\exists A \in \mathcal F$ and $x \in A$. Because $A \in (\mathcal F \cup \{B\} )$, then definitely $x \in \bigcup(\mathcal F \cup \{B\})$
Now suppose $x \in \bigcup(\mathcal F \cup \{B\})$, then $\exists C \in \mathcal F$, such that $x \in C$ or $x \in B$. So there are two cases:
Case 1. $\exists C \in \mathcal F$, such that $x \in C$, then $x \in \bigcup F$, then $x \in B \cup (\bigcup \mathcal F)$.
Case 2. $x \in B$, then $x \in B \cup (\bigcup \mathcal F)$ $\blacksquare$