Prove or disprove the following proposition $L_1^*∪L_2^*⊆(L_1∪L_2)^*$ $L_1^*∪L_2^*⊆(L_1∪L_2)^*$

I actually disproved the opposite proposition $[(L_1∪L_2)^*⊆L_1^*∪L_2^*]$ and my intuition tells me that this is actually true... I tried to show that the combinations of words in the right group $[(L_1∪L_2)^*]$ obviously contain all the words from the left group $[L_1^*∪L_2^*]$ and it's also very plain to see, but I have no idea how to formalize it.

Any idea how to formally prove it?
Thanks.
This is true. 
You only have to prove the following staments:


if $A\subseteq B$ then $A^{*} \subseteq B^{*}$,
If $A\subseteq C$ and $B\subseteq C$ then $A \cup B \subseteq C$.


We know $L_1 \subseteq L_1 \cup L_2$ for 1. we can conclude $L_1^{*} \subseteq (L_1 \cup L_2)^{*}$, analogously $L_2^{*} \subseteq (L_1 \cup L_2)^{*}$. For 2. we can conclude $L_1^{*} \cup L_2^{*} \subseteq (L_1 \cup L_2)^{*}$.