Prove that if $mathcal{F} subseteq mathcal{G}$ then $cupmathcal{F} subseteq cupmathcal{G}$

Suppose $mathcal{F}$ and $mathcal{G}$ are families of sets. Prove that if $mathcal{F} subseteq mathcal{G}$ then $cupmathcal{F} subseteq cupmathcal{G}$

My attempt:

Given $mathcal{F} subseteq mathcal{G}$, writing out logical form of Goal we have
$$exists A in F(x in A) rightarrow exists A in G(x in A)$$

Now assuming $exists A in F(x in A)$ (putting it to list of givens), our Goal is now to obtain

$$exists A in G(x in A)$$

Now, applying Existensial Instantiation, let us assume there exists $A_{0}$, such that $A_{0}$ $in F$. So now $x in A_{0}$
I am feeling stuck here.Thanks

EDIT