Can topological spaces arising as suspensions be characterised this way? Let $X$ be a topological space. The suspension of $X$ is the topological space $SX = X\times[0,1]/\sim$ where $(x_1, 0) \sim (x_2, 0)$ and $(x_1, 1)\sim (x_2, 1)$ for all $x_1, x_2 \in X$. 
Let $p : X\times I \to SX$ be the projection. Note that $p(X\times[0, 1))$ and $p(X\times(0, 1])$ are open contractible subsets of $SX$ and their intersection is homotopy equivalent to $X$. Is the existence of such sets enough to characterise topological spaces which arise as suspensions? That is,

If $Y$ is a connected topological space and $U, V \subset Y$ are open contractible subsets with $U\cup V = Y$ and $U\cap V$ homotopy equivalent to $X$, is $Y$ homotopy equivalent to $SX$?

 A: If you want strong homotopy equivalence then the answer is no. The pseudocircle is a finite topological space with four points $X = \{a,b,c,d\}$ and open sets $X$, $\{a,b,c\}$, $\{a,b,d\}$, $\{a,b\}$, $\{a\}$, $\{b\}$, and $\varnothing$. Then $X$ is the union of the two contractible open sets $\{a,b,c\}$ and $\{a,b,d\}$ whose intersection is $\{a,b\}$ with the discrete topology, i.e. $S^0$. But $X$ is not homotopy equivalent to the suspension of $S^0$, AKA the circle. It is however weakly equivalent to it.
A: On p.247 of Topology and Groupoids (T&G) we have 
7.4.3 (Corollary) Let the space $X$ be the union of closed subspaces $X_1, X_2$ with intersection $X_0$ such that the inclusions $X_0 \to X_1, X_0 \to X_2$ are closed cofibrations. Suppose that $X_1$ and $X_2$ are contractible. Then $X$ is of the homotopy type of the suspension $SX_0$. 
This is a Corollary of 7.4.3 (The gluing theorem), which first appeared in the 1968 (differently titled) edition of that book. I thought then, and still think,  that students should learn how to construct homotopy equivalences before they learned how  to show such did not exist. 
See also this mathoverflow discussion. 
November 15: Actually there is a more precise result. Since $X_1$ is contractible, the identity map $X_0 \to X_0$ extends to a map of pairs $f:(CX_0,X_0) \to (X_1,X_0)$, where $CX_0$ is the cone on $X_0$. But the map $CX_0 \to X_1$ is a homotopy equivalence, since both are contractible. Now we can use 7.4.2 of T&G, and that $X_0 \to X_1$ is a cofibration,  to get that $f$ is a homotopy equivalence of pairs. 
