I'm trying to prove this lemma:
$\begin{align} &\text{Let }A\text{ in }B\text{ be subsets of }V(\Gamma)\text{ for some graph }\Gamma.\\ &\text{If we define }\partial(A)\text{ to be the set of all edges of graph }\Gamma\\ &\text{with one end in }A\text{ and one not in }A\text{, then}\\ &|\partial(A\cup B)|+|\partial(A\cap B)|\le|\partial A|+|\partial B|. \end{align}$
It's supposed to be an easy proof, but I don't see it. All I manage to get is that $|\partial(A\cup B)|+|\partial(A\cap B)|\le 2\cdot(|\partial A|+|\partial B|)$.
I also have a hint, but I don't know how to use it.
Hint: the difference between the two sides is twice the number of edges joining $\;A\setminus B\;$ to $\;B\setminus A\;$.
This lemma will help me prove the lemma from this question: Proving that distinct edge atoms of a graph are vertex-disjoint. And with that lemma I can prove that every vertex-transitive graph has edge connectivity equal to its valency.
For disjoint subsets $C,D \subseteq V$, let $e(C,D)$ denote the number of edges with one endpoint in $C$ and the other in $D$. Also, for clarity, let $X$ denote $V - (A \cup B)$.
Each edge in $\partial A$ has one endpoint in either $A - B$ or $A \cap B$, and the other endpoint in either $B - A$ or $X$. This gives the following identity: $$|\partial A| = e(A - B, B - A) + e(A - B, X) + e(A\cap B, B - A) + e(A \cap B, X) .$$
Analogously, we get $$|\partial B| = e(B - A, A - B) + e(B - A, X) + e(A\cap B, A - B) + e(A \cap B, X).$$
Note also that
$$|\partial (A \cup B)| = e(A - B, X) + e(B - A, X) + e (A \cap B, X)$$
and
$$|\partial (A \cap B)| = e (A\cap B, A - B) + e(A \cap B, B - A) + e(A \cap B, X).$$
Now if you can avoid getting lost in this jumble of letters, by adding the first two identities you can verify that
$$|\partial A| + |\partial B| = 2 e(A - B, B - A) + |\partial (A \cup B)| + |\partial (A \cap B)|,$$
which implies the inequality in the title.