Let $F = \{S_1,\dots,S_m\}$ and $G = \{T_1,\dots,T_m\}$ be two collections of subsets of a finite set $E$.
A transversal for $F$ is a list of elements $s_1,\dots,s_m$, one coming from each set in $F$. A common transversal is a single transversal for both $F$ and $G$.
I am trying to understand the following theorem from page 120 of "Introduction to Graph Theory, 4th ed" by Robin J. Wilson:
Theorem: $F$ and $G$ have a common transversal if and only if, for all subsets $A$ and $B$ of $\{1,\dots,m\}$, $$ | ( \bigcup_{i \in A} S_i ) \cap (\bigcup_{j \in B} T_j ) | \geq |A| + |B| - m. $$
My question: A sketch of the proof is given, but I do not understand what it is saying and/or there may be a typos. Can you illustrate what the proof is saying with an example and/or correct it? Here is the verbatim proof in its entirety:
Proof: Consider the family $X= \{X_i\}$ of subsets of $E \cup \{1,\dots,m\}$, assuming that $E$ and $\{1,\dots,m\}$ are disjoint, where the indexing set is also $E \cup \{1,\dots,m\}$ and where $X_i = S_i$ if $i \in \{1,\dots,m\}$ and $X_i = \{i\} \cup \{j: j \in T_j\}$ if $i \in E$. It is not difficult to verify that $F$ and $G$ have a common transversal if and only if $X$ has a transversal. The result follows by applying Hall's marriage theorem on the family $X$.