Open simply connected covering implies simply connectedness of the space Let $\{U_i\}$ be an open covering of the space $X$ having the following properties: 


*

*There exists a point $x_0$ in $U_i$ for all $i$.

*Each $U_i$ is simply connected.

*If $i$ and $j$ are different, then the intersection of $U_i$ and $U_j$ is path connected. 


Now, I want to prove that $X$ is simply connected. I can prove it separately by Lebesgue Number Lemma, Compactness of the loop in $X$, and Van Kampen theorem. Can you please recommend me another solution to this problem? I think there should be some cool and straightforward solutions. Thank you.
Edit: I just noticed that one of my solutions was not correct. I used Van Kampen theorem incorrectly. I assumed the intersection of any three sets is path-connected, which is not essentially true.
 A: @diracdeltafunk is correct: the thing you're trying to prove is false. 
Consider a figure-$8$ lying on its side. By chopping off the rightmost semicircle, you get something that looks like the letter $\alpha$. Let's call that $U_1$. Now chop off the left-most semicircle similarly, and get $U_2$. The crossing point is in both; the intersection looks like the letter $X$, which is certainly simply connected, but the figure-8 is certainly not simply connected. 
A: Unless I am misreading your question this remains untrue. Cover $S^1$ with 3 open intervals.
A: Here's a sketch of a pretty direct solution. It sounds like what you already have. I doubt there's anything simpler.
Suppose we have a loop $\alpha$ at $x_0$. We write this as a finite sequence of subpaths where each subpath lies in some $U_i$. (I think this is where Lebesgue's Number Lemma comes in.) At the end of each subpath, insert a path to $x_0$ and back, using the path-connectedness of the intersection of the two $U_i$. This gives us a loop $\alpha'$, homotopic to $\alpha$, which is a product of loops $\beta_j$ each lying within one of the $U_i$. Each $\beta_j$ is nullhomotopic, and therefore so is $\alpha'$, and $\alpha$.
