Show $X$ is simply-connected given properties of two subsets I'm given: 
$X$ is a manifold, $X = U\cup V$ for $U,V \subset X$ open, connected and simply-connected, with $U \cap V$ connected.
And given this, I want to show $X$ is simply connected.
Attempt
I can see why all the hypotheses are necessary, by considering for example, two sausages in the plane which intersect in two separate discs and unite for an annulus. Also, if $U \cap V = \emptyset$ I can prove it. Otherwise, all I've thought of is to take a loop $p : [0,1] \rightarrow X $ and split $[0,1]$ into countably-many subintervals $I_i$ whose images are contained in open sets $A_i$ homeomorphic to $\Bbb{R}^n$ (for $n =$ dim$X$), and perhaps use convexity of the $A_i$...to try and homotope $p$ to a null path.
Would really appreciate some help, or a complete solution, as topology puzzles tend to beat me; and if I saw a solution to this it might get me thinking on the right track for future ones. It's exercise 3.1 in Forster's "Lectures on Riemann Surfaces", slightly edited to add in the assumption $X = U \cup V$ as without that I think it fails (take $U = V$ to be a disc within an annulus $X$ !)
Edit I'm sure there was an answer yesterday explaining how it follows from the Van-Kampfen theorem, but it seems to be gone now; thank you to whoever posted it. 
 A: Pick a basepoint $x_0 \in U\cap V$. Let $p:I \to X$ be some loop starting at $x_0$. Then we can write $p$ as a finite product of paths $p = f_1 \cdot \ldots \cdot f_n$, where each $f_i$ is entirely contained in either $U$ or $V$. WLOG we can assume that each $f_{2i}$ is entirely contained in $U$ and each $f_{2i+1}$ is entirely contained in $V$ (so they alternate). Then all endpoints of these paths are contained in $U\cap V$.
For each $1\leq i < n$, let $g_i$ be a path in $U\cap V$ from $f_i(1)=f_{i+1}(0)$ to $x_0$. Such a path always exists since $U\cap V$ is connected. Then we have that $$p = f_1 \cdot g_1 \cdot \overline{g_1} \cdot f_2 \cdot \ldots \cdot g_{n-1} \cdot \overline{g_{n-1}} \cdot f_n$$ Note that $(f_1 \cdot g_1)$, $(\overline{g_{n-1}}\cdot f_n)$, and $(\overline{g_i}\cdot f_{i+1} \cdot g_{i+1})$ for each $i$ are all loops based at $x_0$ which are contained entirely in either $U$ or $V$. Since $U$ and $V$ are simply connected, each of these loops is null-homotopic,  hence their product, $p$, is too.
