Proving that distinct edge atoms of a graph are vertex-disjoint. The Question
In Godsil and Royle's Algebraic Graph Theory, they prove that any two distinct edge atoms of a graph $X$ are vertex-disjoint, making the following argument, where $A$ and $B$ are such edge atoms:

So we may assume that $A\cup B$ is a proper subset of $V(X)$. Now, the previous lemma yields
$$|\partial(A\cup B)| + |\partial(A\cap B)|\le 2\kappa_1(X),$$
and, since $A\cup B\ne V(X)$ and $A\cap B\ne\emptyset$, this implies that
$$|\partial(A\cup B)| = |\partial(A\cap B)| = \kappa_1(X).$$

I don't understand why the second assertion is true, nor why it follows from  $A\cup B\ne V(X)$ and $A\cap B\ne\emptyset$.
Background
I'm unsure how common their notation/definitions are, so I've included some below.

*

*We let $\kappa_1(X)$ denote edge-connectivity;


*For $S\subseteq V(X)$, we define $\partial S$ to be $\{xy\in E(X):x\in S, y\notin S\}$. That is, it is the set of edges with one end in $S$ and one end not in $S$;


*An edge atom of $X$ is a subset $S\subseteq V(X)$ such that $|\partial S|=\kappa_1(X)$, and $|S|$ is minimal.


*The lemma referenced states that $|\partial(A\cup B)|+|\partial(A\cap B)|\le |\partial A| + |\partial B|$ for $A,B\subseteq V(X)$.
 A: The second assertion follows from the first because, by Menger's Theorem, for any nonempty $S \subsetneq V(X)$, $|\partial S| \geq \kappa_1(X)$.
Here's another way to think about it. If either of $\partial (A \cup B)$ or $\partial (A \cap B)$ had size different from $\kappa_1 (X)$, then one of the two would have to be less than $\kappa_1 (X)$, which contradicts $\kappa_1 (X)$ being the edge connectivity, since neither $A \cap B$ or $A \cup B$ is $V(X)$ or empty.
A: In Godsill & Royle's book :
"So we may assume that A∪B is a proper subset of V(X)."
This is proved immediately above the statement on page 38. So A∪B≠V(X) since it is a proper subset (and is also not the empty set since neither A nor B are empty).
"A∩B≠∅" should be stated in the proof as an assumption because the object of corollary 3.3.2 is to prove A and B are disjoint.
The final line:
" |∂(A∪B)|=|∂(A∩B)|=κ1(X)."
follows the expression from the previous lemma because neither |∂(A∪B)| nor |∂(A∩B)| can be less than κ1 (and both are non-zero from the previous working) since the edge connectivity is κ1 so the only option is  that both equal κ1
So the corollary is proved by contradiction.
