Suppose $X_1,X_2$ form a closed cover of $X$ and the $U \cap X_i$ are open. Is $U$ open? Equivalently, is $X$ the pushout of the $X_i$ over their intersection?
If $X$ has a finite closed cover of subspaces $X_1$ and $X_2$ then $F \subset X$ is closed iff $F \cap X_1$ is closed in $X_1$ and $F \cap X_2$ is closed in $X_2$. One direction uses the definition of the subspace topology (left to right) and right to left uses 2 facts: firstly, a subset $A$ that is closed (in the subspace topology) in $C$ and $C$ is closed in $X$, then $A$ is closed in $X$, and secondly that a union of 2 closed subsets of $X$ is again closed in $X$.  We apply these facts to $F \cap X_1$ and $F \cap X_2$, whose union is again $F$ as the $X_i$ form a cover of $X$.

Now, we also have, dually, that $O$ is open in $X$ iff $O \cap X_1$ is open in $X_1$ and $O \cap X_2$ is open in $X_2$. This holds as 

$O$ open in $X$ 

iff $X\setminus O$ closed in $X$ 

iff $(X \setminus O) \cap X_1$ closed in $X_1$ and $(X \setminus O) \cap X_2$ closed in $X_2$ 

iff $X_1 \setminus ( (X \setminus O) \cap X_1 ) = O \cap X_1$ open in $X_1$ and $X_2 \setminus (( X \setminus O) \cap X_2) = O \cap X_2$ open in $X_2$.
The union of two open sets is open...