Topological Space $X$ as closed union 
Let $X$ be a topological space. Suposse $X=X_1\cup X_2$ where $X_1, X_2$ are closed. Prove that if $B\subset X_1$ and $B\cap X_2\subset A$ where $A$  is open in $X_2$ then $B\subset \operatorname{Int} ( X_1\cup A)$.

I have tried a lot of ways to prove this. But the part in strong text I think is suspect. Because the open set is in $X_2$ not in $X$. But I did not find any counterexample. 
I appreciate any help to this problem. 

First, let $U \subseteq X$ be open such that $U \cap X_2 = A$. Note that it must be that $U \subseteq X_1 \cup A$. Now consider the following partition of $B$: $( B \cap X_2 ) \cup ( B \setminus X_2 )$.


*

*By hypothesis, $B \cap X_2 \subseteq A \subseteq U \subseteq X_1 \cup A$, and since $U$ os open it must be that $B \cap X_2 \subseteq \mathrm{Int} ( X_1 \cup A )$.

*Note that $B \setminus X_2 \subseteq X \setminus X_2 \subseteq X_1 \subseteq X_1 \cup A$, and since $X \setminus X_2$ is open it follows that $B \setminus X_2 \subseteq \mathrm{Int} ( X_1 \cup A)$.
Therefore $B \subseteq \mathrm{Int} ( X_1 \cup A )$.
