If L1 ⊆ L2 and L2 is regular, then L2 − L1 is regular
You can show that if $L_1 \subseteq L_2$ and $L_2$ is regular, then $L_2 \setminus L_1$ is regular iff $L_1$ is regular. This follows from the product construction, for example. Taking $L_2 = \Sigma^*$, we find that $\Sigma^* \setminus L_1$ is regular iff $L_1$ is regular. Hence if $L_1$ is any non-regular language, then it refutes your statement taken with $L_2 = \Sigma^*$.