Be $G$ a group with subgroups $U$ and $V$ and be $UV := \left\{uv \mid u \in U, v \in V \right\}$
I want to prove the following:
$UV$ is in general no subgroup of $G$.
Proof. Be $U \nsubseteq V$ and $V \nsubseteq U$. Be $x \in U \setminus V$ and $y \in V \setminus U$.
Hence $xy \in UV$ but $xy \notin U \cup V$. This implies $\left(xy \right)\left(xy\right) \notin UV$ per definition of $UV$, hence $UV$ is no subgroup of $G \quad \square$
But I also want to give an explicit counterexample beside this proof. Could you give me a hint in which group I should look for the easiest counterexample?
Let us take $S_3$ the symmetric group on three labels.
If $S_3=\{1\!\!1,(12),(13),(23),(123),(321)\}$ then take $U=\{1\!\!1,(12)\}$ and $V=\{1\!\!1,(13)\}$. They are subgroups.
You will see that $UV$ is not a subgroup.