Question: Let $X = X_1 \cup ... \cup X_n$ be a subset of $\mathbb{R}^m$ such that $X_i$ is convex and $X_i \cap X_j \cap X_k \neq \emptyset$ for all $i, j, k$, then $\pi_1(X, x_0) = \{e\}$ for all $x_0 \in X$.
My thoughts: all $X_i$ are path connected, and for $x \in X_i, y \in X_j$ they have a non-empty intersection, so pick a point $z$ in that, and then we can construct a path $x \to z \to y$. This means that $X$ is path connected, so no need to worry about basepoints.
We know that $X_1 \cup X_2 \cup X_3$ has a trivial fundamental group because it is contractible (take $a \in X_1 \cap X_2 \cap X_3$, then $H_t(x) = tx + (1 - t)a$ is a deformation retract). I don't know how to extend this to $X$ entire. What is the significance of requiring $X_i \cap X_j \cap X_k \neq \emptyset$? I've encountered $X_i \cap X_j \neq \emptyset$ lots of times, but never with 3.