Question: Equivalently, is $X$ the pushout of the $X\_i$ over their intersection?

Answer: 
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...