The fact is true. Here is the proof that if $A\cup B$ is normal in $G$, then $A\cap B$ is normal in $G$ or $A^g$ is inside $B$ for some $g$. Suppose that $A^g$ is not inside $B$ for any $g$. If $A\cup B$ is normal in $G$, then $A$ and $B$ are normalized by an index at most 2 subgroup $G_1$ of $G$. Indeed, for every $g\in G$, we have $A^g$ is a union of two subgroups $A^g\cap A$ and $A^g\cap B$. That can happen only if $A^g=A$ or $A^g=B$, same for $B$.
Now if $G_1=G$, there is nothing to prove. If $G_1$ is of index 2, then there exists an element $g$ with $A^g=B$, $B^g=A$ and $G=\langle G_1, g\rangle$. Therefore $A\cap B$ is normal in $G$. In particular, it is normal in $A$ and $B$. Now suppose that $A^g$ is inside $B$ for some $g$. Notice that for every $h\in G$ either $A^h=A$ or $A^h\subseteq B$ and similarly either $B^h=B$ or $B^h\subseteq A$. If $A^h\subseteq B$ and $B^h \subseteq A$ then $(A\cap B)^h=A\cap B$. Therefore the normalizer $C$ of $A\cap B$ together with the normalizer $D$ of $B$ gives the whole $G$. This again implies that either $C=G$ and $A\cap B$ is normal in $G$ or $D=G$, and $B$ is normal in $G$. Thus the only case left is when $A^g\subseteq B$ for some $g$ and $B$ is normal in $G$. In that case $A\subseteq B^{g^{-1}}=B$, so $A\cap B=A$ and since it must be of finite index in $B$ by assumption, we conclude that $A=A\cap B$ has a normal subgroup of finite index both in $A$ and in $B$.