Is $d(A, B) =|(A\Delta B)|$ a metric on all finite subsests? Let $F(S) $ be the set of all finite subsets of a set
$S$. For all
$A, B ∈ F(S)$,
let $∆
(A, B)=(A\setminus B) ∪ (B\setminus A) $ be the symmetric difference
between
A
and B. Let $d(A, B
) $ be the cardinality of
$∆
(A, B).$ Is
$d$
a metric? 
I am able to prove positivity, definiteness, symmetry. For any three sets $A, B, C$ I have to show $|(A\Delta B)| \le |(A \Delta C) |+| (C\Delta B) |$.
My intuition suggests triangle inequality may also hold, but I can't make it rigorous, please elaborate.

Since symmetric difference is commutative and associative, we have
$$A\Delta B\ =\ (A\Delta C)\,\Delta\, (C\Delta B)\ \subseteq (A\Delta C)\cup(C\Delta B)$$
and the latter has at most $|A\Delta C|+|C\Delta B|$ elements.
