Open simply connected covering implies simply connectedness of the space

Question

Let $\{U_i\}$ be an open covering of the space $X$ having the following properties:

1. There exists a point $x_0$ in $U_i$ for all $i$.
2. Each $U_i$ is simply connected.
3. 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.

Answer 1

@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.