I don’t quite understand why the empty set is not considered in the proof of the theorem.
The proof I have come across is:
Suppose $\cup \mathcal{F} \subseteq B$. Let $x$ be an arbitrary element of $\mathcal{F}$. Let $y$ be an arbitrary element of $x$. Since $y \in x$ and $x \in \mathcal{F}$, clearly $y \in \cup \mathcal{F}$. But then since $\cup \mathcal{F} \subseteq B$, $y \in B$. Since $y$ was an arbitrary element of $x$, we can conclude that $x \subseteq B$, so $x\in \mathbb{P} (B)$. But $x$ was an arbitrary element of $\mathcal{F}$, so this shows that $\mathcal{F} \subseteq \mathbb{P} (B)$, as required.
The main question I have is with the statement: ‘Let $y$ be an arbitrary element of $x$’ implies that the set $x$ is non-empty but isn’t $\emptyset \in \mathcal{F}$?
The other question is regarding the notation of sets. Throughout the book sets have been indicated by capital letters and elements of sets using lower case letters. Why would the author use $x$ to represent an arbitrary set (element) of $\mathcal{F}$? Are there general guidelines for choosing variables?
I'd write the same proof as follows :
Let $F$ be any element of $\mathcal{F}$.
[it makes more sense expositionally to have sets be capitals, families of sets use
\mathcalfont etc., elements small letters etc. But that's just convention.]If $F = \emptyset$, $F \in \Bbb P(B)$ trivially and we're done. [this step is not strictly necessary, see below, but could be added]
Let $x \in F$ be arbitrary, then $x \in \bigcup \mathcal{F} \subseteq B$ so $x \in B$ and hence $F \subseteq B$.
[This would also have worked for the empty set, which is a subset of any set we like. Void truth etc.]
Hence $F \in \Bbb P(B)$ and the inclusion has been shown.
But be flexible in what you accept as notations, the proof you gave using $x$ as element of $\mathcal{F}$ might be confusing at first, but in set theory everything is a set, so distinguishing different levels of abstraction is not as important.