I am trying of resolve this exercise: Show that if $|\psi \rangle$ is an entangled state of two Qbits, then the application of a unitary operator of the form $U_1 \otimes U_2$ necessarily generates an entangled state.
Suppose that $(U_1 \otimes U_2)|\psi\rangle$ is not entangled then it must have the form
$$(U_1 \otimes U_2)|\psi\rangle = (a|0\rangle+b|1\rangle)\otimes (c|0\rangle+d|1\rangle),$$ but the unique way this will be true is when $|\psi\rangle$ is not entangled due to definition of linear operator.
My proof is bad? Help me please