Exercise 2 in Hatcher, section 1.2: the union of convex sets is simply connected Here is my problem : 

I have convex subsets $X_1, \dots, X_n$ of $\mathbb R^m$ such that $X_i \cap X_j \cap X_k$ is never empty. I have to show that $\bigcup\limits_{i=1}^nX_i$ is simply connected.

This is the chapter about Van Kampen's theorem so I thought we have to use Van Kampen's theorem. Almost all the hypothesis are true.
But I just read that the intersection $\bigcap\limits_{i=1}^nX_i$ is not always non-empty, and a inside this intersection is needed for applying Van Kampen's. 
Did I make some mistakes? Maybe there is a smarter way to apply Van Kampen's Theorem here?
Thanks.
 A: I think the fact that the cover is finite is not really needed here. Let's change $X_1,...,X_n$ to $\{X_i\}$. Let $i_0$ be a fixed index and $x_0\in X_{i_0}$. Define $Y_i=X_{i_0}\cup X_i$. Note that $Y_i$ is open path connected because $X_{i_0}$ and $X_i$ are path connected open sets with nonempty intersection. $Y_i\cap Y_j=X_{i_0}\cup(X_i\cap X_j)$ which is path connected because $X_{i_0}\cap X_i\cap X_j\neq \emptyset$ and both $X_{i_0}$ and $X_i\cap X_j$ are path connected. Each $Y_i$ is also simply connected by applying van Kampen to $\{X_i,X_{i_0}\}$. After checking all of these, we can apply van Kampen to $\{Y_i\}$ and thus conclude that $X$ is simply connected.
A: Do it by induction on $n$. The case $n = 2$ is exactly Van Kampen's theorem: $X_1$ and $X_2$ are path-connected, and so is their intersection (it's the intersection of two convex sets, hence convex, and it's nonempty, hence path-connected). Thus
$$\pi_1(X_1 \cup X_2) \cong \pi_1(X_1) *_{\pi_1(X_1 \cap X_2)} \pi_1(X_2) = 0.$$
Now for the induction step, suppose that you know the result is true for a given $n \ge 2$ and let $X_1, \dots, X_{n+1}$ be convex sets such that $X_i \cap X_j \cap X_k$ is connected for all $i,j,k$. By the induction hypothesis, $Y = X_1 \cup \dots \cup X_n$ is simply connected. It remains to show that $Y \cap X_{n+1}$ is path-connected, and you can apply Van Kampen's theorem again to conclude. (1)
So let's show $Y \cap X_{n+1}$ is path-connected. Clearly
$$Y \cap X_{n+1} = (X_1 \cap X_{n+1}) \cup \dots \cup (X_n \cap X_{n+1}).$$
Now suppose $x,y$ belong to $Y$, we are looking for a path from $x$ to $y$. Let
$$x \in X_i \cap X_{n+1}, \qquad y \in X_j \cap X_{n+1}.$$ By the hypothesis on the $X_\cdot$, the intersection $X_i \cap X_j \cap X_{n+1}$ is non-empty; choose $z$ inside it. Since $X_i \cap X_{n+1}$ is path-connected (it's the nonempty intersection of two convex sets), there is a path $\gamma$ from $x$ to $z$. Similarly, there is a path $\gamma'$ from $z$ to $y$. Concatenating these two path gives a path $\alpha = \gamma \cdot \gamma'$ from $x$ to $y$. Thus $Y \cap X_{n+1}$ is path-connected and we can conclude (cf. (1)).
Remark. You forgot the assumption (which is written in Hatcher's book) that the sets $X_i$ have to be open. This is crucial to apply van Kampen's theorem, in general it's false. 
