Definition 1: An algebra of sets on a non-empty set $X$ is a non-empty collection $\cal{A}$ of subsets of $X$ that is closed under taking complements and finite unions.
Definition 2: An elementary family of sets on a non-empty set $X$ is a collection $\cal{E}$ of subsets of $X$ such that (i) $\emptyset \in \cal{E}$, (ii) $\cal{E}$ is closed under finite intersection, (iii) if $E \in \cal{E}$, then $E^c$ is a finite disjoint union of members of $\cal{E}$.
Proposition: If $\cal{E}$ is an elementary of sets, then the collection $\cal{A}$ of finite disjoint unions of members of $\cal{E}$ is an algebra of sets.
Proof: It's easy to show that if $A,B \in \cal{E}$, then $A \setminus B \in \cal{A}$ (just write $A \setminus B = A \cap B^c$ and apply property (iii) to $B^c$). Using this observation, we write $A \cup B = (A \setminus B) \sqcup B$, and conclude that $A \cup B \in \cal{A}$, as well, for any $A,B \in \cal{E}$. The rest of the proof follows by induction.
Question: Is every elementary family of sets in fact an algebra of sets?
Look at the proof: we make an observation there that if $A,B \in \cal{E}$, then $A \cup B \in \cal{A}$. But $\emptyset \in \cal{E}$, so we may as well say that if $A \in \cal{E}$, then $A = A \cup \emptyset \in \cal{A}$.
Reference: Folland, Real Analysis, pp. 23-24.