$\Delta(\mathbf{U}\cdot\mathbf{V})=\nabla\cdot\nabla(\mathbf{U}\cdot\mathbf{V})$, since $(\mathbf{U}\cdot\mathbf{V})$ is a scalar.
Now, use the relation $\nabla(\mathbf{A}\cdot\mathbf{B})=(\nabla\mathbf{A})\cdot\mathbf{B} + (\nabla\mathbf{B})\cdot\mathbf{A}$ and get
$\Delta(\mathbf{U}\cdot\mathbf{V})=\nabla\cdot[(\nabla\mathbf{U})\cdot\mathbf{V} + (\nabla\mathbf{V})\cdot\mathbf{U}]$
Now, use the relation $(\mathbf{A}\mathbf{B})\cdot\mathbf{C}=\mathbf{A}(\mathbf{B}\cdot\mathbf{C})$ and get
$\Delta(\mathbf{U}\cdot\mathbf{V})=\nabla\cdot[\nabla(\mathbf{U}\cdot\mathbf{V}) + \nabla(\mathbf{V}\cdot\mathbf{U})]$, or
$\Delta(\mathbf{U}\cdot\mathbf{V})=\nabla\cdot[2\nabla(\mathbf{U}\cdot\mathbf{V})]=2\nabla\cdot\nabla(\mathbf{U}\cdot\mathbf{V})=2\Delta(\mathbf{U}\cdot\mathbf{V})$.
Then
$\Delta(\mathbf{U}\cdot\mathbf{V})=2\Delta(\mathbf{U}\cdot\mathbf{V})$, leading to
$\Delta(\mathbf{U}\cdot\mathbf{V})=0$
Is it correct?
(1) $\Delta(\mathbf{U}\cdot\mathbf{V})=\nabla\cdot\nabla(\mathbf{U}\cdot\mathbf{V})$ is true, checked
(2) $\nabla(\mathbf{U}\cdot\mathbf{V})=(\nabla\mathbf{U})\cdot\mathbf{V}+(\nabla\mathbf{V})\cdot\mathbf{U}$ is also true, derived
(3) $(\mathbf{U}\mathbf{V})\cdot\mathbf{W}=\mathbf{U}(\mathbf{V}\cdot\mathbf{W})$ is also true, easily derived
What is not true is $(\nabla\mathbf{U})\cdot\mathbf{V}=\nabla(\mathbf{U}\cdot\mathbf{V})$. In the Lhs the nabla is acting upon U only, while in the Rhs it is acting upon the dot product of both U and V. Checked a case and (3) may hold for vector fields but it does not hold when nabla is part of it.
Naturally then it is not true that $\Delta(\mathbf{U}\cdot\mathbf{V})=0$ or that $\nabla(\mathbf{U}\cdot\mathbf{V})=0$ These do not follow from the above general case