Lemma. Let $X$ be a topological space, and let $A\subseteq X$. Then $\mathrm{int}(A) = X-\overline{X-A}.$
Proof. Remember that $\mathrm{int}(A)$ is the largest open subset $U$ of $X$ such that $U\subseteq A$, and that $\overline{B}$ is the smallest closed subset $C$ of $X$ such that $B\subseteq C$.
If $C$ is a closed set such that $X-A\subseteq C$, then $X-A\subseteq \overline{X-A}\subseteq C$, hence taking complements we have $X-C\subseteq X-\overline{X-A}\subseteq X-(X-A) = A$. Since $C$ is closed, $X-C$ is open, so $X-C$ is an open set contained in $A$; hence $X-C\subseteq \mathrm{int}(A)$. In particular, $X-\overline{X-A}\subseteq \mathrm{int}(A)$.
Now let $U$ be an open set contained in $A$; then $X-U$ is a closed set that contains $X-A$; since the closure is the smallest closed subset that contains the set, then $\overline{X-A}\subseteq X-U$. So $U=X-(X-U) \subseteq X-\overline{X-A}$. In particular, $\mathrm{int}(A)\subseteq X-\overline{X-A}$. This proves the desired equality. QED
Corollary. $X-\mathrm{int}(A) = \overline{X-A}$.
So we have that $\begin{align*} \mathrm{int}(F\cup \mathrm{int}(A)) &= X - \overline{X-(F\cup\mathrm{int}(A))}\\ &= X - \overline{(X-F) \cap (X-\mathrm{int}(A))}\\ &= X - \overline{(X-F) \cap \overline{X-A}}\\ &= X - \overline{(X-F)\cap (X-A)}&\qquad&\text{(using your fact)}\\ &= X - \overline{(X-(F\cup A))}\\ &= \mathrm{int}(F\cup A). \end{align*}$ as desired.