Hatcher's Algebraic Topology and Star-Shaped Neighbourhoods Questions on this particular question (1.1.4) from Hatcher's Algebraic Topology have been asked before on Math Stack Exchange. Alas, I find nothing that properly answers the part that I'm currently tearing my hair out concerning.

A subspace  $X\subset \Bbb R^n$ is said to be star-shaped if there is a point $x_0 \in X$ such that, for each $x \in X$, the line segment from $x_0$ to $x$ lies in $X$.
Show that if a subspace $X \subset \Bbb R^n$ is locally star-shaped, in the sense that every point of $X$ has a star-shaped neighborhood in $X$, then every path in $X$ is homotopic in $X$ to a piecewise linear path, that is, a path consisting of a finite number of straight line segments traversed at constant speed. Show this applies in particular when $X$ is open or when $X$ is a union of finitely many closed convex sets.

It's the part in bold that troubles me. If I am interpreting what Hatcher is saying correctly, then that part can accurately be reformulated as:

Show that if $X \subset \Bbb R^n$ is open or a union of finitely many closed convex sets, then $X$ is locally star-shaped.

This I am however constantly finding myself unable to do.
Any takers?
 A: As Ted Shifrin said in his comment, open balls are star-shaped which covers the case that $X$ is open.
The case that $X$ is a union of finitely many closed convex sets requires more care.
As you quote in your question

A subspace  $X\subset \Bbb R^n$ is star-shaped if there exists a point $x_0 \in X$ such that for each $x \in X$ the line segment from $x_0$ to $x$ lies in $X$.

Let us call any point $x_0 \in X$ having this property a star-point of $X$. In general not every point is a star-point. As a simple example let $X \subset \mathbb R^2$ be the union of the two line segments connecting $x_0 = (0,0)$ with $(0,1)$ and with $(1,0)$. Then $x_0$ is the only star-point of $X$. Note, however, that in convex sets each point is a star-point.
Now let us strengthen the concept of locally star-shaped:

A subspace  $X\subset \Bbb R^n$ is called strictly locally star-shaped if each $x_0  \in X$ has a star-shaped neighborhood $N(x_0)$ in $X$ such that $x_0$ is a star-point for $N(x_0)$.

Clearly "strictly locally star-shaped" implies "locally star-shaped".
Let us observe that

*

*Convex subsets  $X \subset \mathbb R^n$ are strictly locally star-shaped. In fact, we may take $N(x_0) = X$ for each $x_0 \in X$.

Let us now prove the following lemma:

Let $A, B$ be a closed strictly locally star-shaped sets. Then $X = A \cup B$ is strictly locally star-shaped.

This implies, via induction, that a union of finitely many closed strictly locally star-shaped sets is strictly locally star-shaped. By observation  1. we see that a union of finitely many closed convex sets is strictly locally star-shaped and in particular locally star-shaped.
Let us prove the lemma. We start with some observations. Let $U_r(x_0)$ denote the open ball in $\mathbb R^n$ with radius $r$ and center $x_0$.


*The intersection and the union of star-shaped sets with the same star-point $x_0$ is a star-shaped set with star-point $x_0$.


*$X$ is strictly locally star-shaped if and only if each $x_0 \in X$ admits $r > 0$ such that $U_r(x_0) \cap X$ is star-shaped with star-point $x_0$. The "if" part is trivial, the "only if" part follows from 2.: If $N$ is a star-shaped neigborhood of $x_0$ in $X$ with star-point $x_0$, then there exists $r > 0$ such that $U_r(x_0) \cap X \subset N$ and $U_r(x_0) \cap X = (U_r(x_0) \cap X) \cap N = U_r(x_0) \cap N$ is star-shaped with star-point $x_0$.
Now let $x_0 \in X  = A \cup B$. If $x_0 \notin B$, then there exists $d > 0$ such that $U_d(x_0) \cap B = \emptyset$ (note that the distance between $x_0$ and $B$ is positive  because $B$ is closed). There exists $r> 0$ such that $U_r(x_0) \cap A$ is star-shaped with star-point $x_0$. Let $r' =  \min(d,r)$. Then $U_{r'}(x_0) \cap X = U_{r'}(x_0) \cap A = (U_r(x_0) \cap A) \cap U_d(x_0)$ is star-shaped with star-point $x_0$. See 1. The case $x_0 \notin A$ is treated similarly. So let finally $x_0 \in A \cap B$. Choose $r_A, r_B > 0$ such $U_{r_A}(x_0) \cap A$ and $U_{r_B}(x_0) \cap B$ are star-shaped with star-point $x_0$. Let $r = \min(r_A,r_B)$. Then $U_r(x_0) \cap A$ and $U_r(x_0) \cap B$ are star-shaped with star-point $x_0$. Thus $U = (U_r(x_0) \cap A) \cup (U_r(x_0) \cap B)$ is star-shaped with star-point $x_0$. But clearly $U = U_r(x_0) \cap X$.
