Let $X$ be a normal space and let $U_{1}$, $U_{2}$ be open subsets of $X$ such that $X= U_{1} \cup U_{2}$. Show that there are open sets $V_{1}$ and $V_{2}$ such that $\overline{V_{1}} \subset U_{1}$, $\overline{V_{2}} \subset U_{2}$ and $X=V_{1} \cup V_{2}$.
What I've tried:
Note $X \setminus U_{2}$ is closed and $U_{1}$ is an open set containing it, hence by normality there is an open set $V_{1}$ such that $X \setminus U_{2} \subset V_{1} \subset \overline{V_{1}} \subset U_{1}$.
Similarly we can find an open set $V_{2}$ such that $X \setminus U_{1} \subset V_{2} \subset \overline{V_{2}} \subset U_{2}$.
Then from here I don't see how to conclude $X=V_{1} \cup V_{2}$. Perhaps this is a wrong approach. Can you please help?