If a subset $W$ of $A \cap B$ is open in both $A$ and $B$, then $W$ is open in $A \cup B$. Let $X$ be a topological space and $A,B$ be subsets of $X$ with $A∪B = X$.
Show that if a subset $W$ of $A∩B$ is open in both $A$ and $B$, then $W$ is open in $X$. Here $A$ and $B$ have subspace topologies of $X$.
Here's I tried. 
By definition of subspace, there exists open $U$ in topology such that $W=U∩(A∩B)$.
But I can't proceed from this and i don't know where can i use the fact that $A∪B = X$
 A: $W$ open in $A$ implies $W=U\cap A$ similarly, $W=V\cap B$ $U,V$ are open subsets of $X$. $W\subset U\cap V$, this implies that $W=(U\cap V)\cap A=(U\cap V)\cap B$. We deduce that $W\cup W=((U\cap V)\cap A))\cup (U\cap V)\cap B)=(U\cap V)\cap (A\cup B)=(U\cap V)\cap X=U\cap V$. We deduce that $W$ is open since the intersection of two open subsets is open.
A: It should follow from definitions.
To be "open" means nothing more or less than to be in some list of sets-- called a topology--, or to be a direct manipulation of some list of sets.  The big master topology here is the space $X = A\cup B$.
So we are told $W$ is open in $A$ and $W$ is open in $B$.  What does that mean?  It means there is an open set $M_a\subset X$ so that $W = M_a \cap A$ and there is an open set $M_b \subset X$ so that $W = M_b \cap B$.
Since $M_a$ and $M_b$ are open in $X$.  $M_a \cap M_b$ is open in $X$. 
It should be clear by Venn Diagram that $M_a \cap M_b = W$. 
But here is a  proof:
Let $U = M_a \cup M_b$ (which is open as it is a union of open sets).  $U \cap (A\cap B) = W$.   $U \cap (A\setminus W) = M_a\setminus W := V_a$.  $U \cap (B\setminus W) = M_b\setminus W := V_b$.
And $U = W \cup V_a \cup V_b$, the union of three disjoint sets.
$W = (W\cup V_a) \cap (W\cup V_b) = M_a \cap M_b$.
which is now shown to be open in $X$.
