The fact that in ${\mathbb R}^n$ each point of a compact convex set is a convex combination of at most $n+1$ extreme points is a theorem of Carathéodory. You can prove this by induction on $n$.
The case $n=0$ is easy. For the induction step, if $K$ is a compact convex set in ${\mathbb R}^{n+1}$ and $x \in K$, choosing some extreme point $y$ of $K$ we have $x = t y + (1-t) z$ where $0 \le t \le 1$ and $z$ is a boundary point of $K$. $K$ has a supporting hyperplane $H$ at $z$, and $H \cap K$ is a compact convex set in the $n$-dimensional space $H$ whose extreme points are extreme points of $K$. So represent $z$ as a convex combination
$z = \sum_{i=1}^{n+1} c_i z_i$ of at most $n+1$ of these extreme points, and
$x = t y + \sum_{i=1}^{n+1} (1-t) c_i z_i$ is a convex combination of at most $n+2$ extreme points of $K$.
$\newcommand{\Co}{\operatorname{Co}}$If you want to prove that
$$
\Co\left( \bigcup_{i=1}^k X_i \right) \subseteq \left\{\sum_{i=1}^k t_i x_i : x_i \in X_i,t_i\geq 0,\sum_{i=1}^k t_i =1\right\}, \tag 1
$$
you should just assume $x$ is an element of the left side and prove that it follows that it's an element of the right side.
For now I'll take $\Co\left( \bigcup_{i=1}^k X_i \right)$ to be defined as the intersection of all convex sets having $\bigcup_{i=1}^k X_i$ as a subset. To show that if $x$ is a member of the left side of $(1)$, then $x$ is a member of the right side of $(1)$ is to show that if $x$ is a member of every convex set having $\bigcup_{i=1}^k X_i$ as a subset, then $x$ is a member of the right side of $(1)$. For that, it is enough to show that the right side of $(1)$ is a convex set having $\bigcup_{i=1}^k X_i$ as a subset. Then we would have
- $x$ is a member of every convex set having $\bigcup_{i=1}^k X_i$ as a subset.
- The right side of $(1)$ is a convex set having $\bigcup_{i=1}^k X_i$ as a subset.
- Therefore, $x$ is a member of the right side of $(1)$.
So the question is: How do we prove the second statement in the bulleted list above?
Notice that if $t_j=1$ and $t_i=0$ for all $i\ne j$, then $t_i\ge0$ for $i=1,\ldots,k$ and $t_1+\cdots+t_k = 1$, so $t_1 x_1 + \cdots + t_k x_k$ is a member of the right side of $(1)$. But $t_1 x_1 + \cdots + t_k x_k = x_j \in X_j$. Thus $x_j$ is a member of the right side of $(1)$. Since $x_j$ could have been any member of $X_j$ at all, we conclude that every member of $X_j$ is a member of the right side of $(1)$, so $X_j$ is a subset of the right side of $(1)$. Since the index $j$ may be any of $1,\ldots,k$, we must conclude that all of $X_1,\ldots,X_k$ are subsets of the right side of $(1)$. Therefore $\bigcup_{i=1}^k X_i$ is a subset of the right side of $(1)$.
Next we must prove that the right side of $(1)$ is convex. So suppose $x$ and $y$ are members of that set; we must prove that every convex combination $cx+(1-c)y$ is a member of the right side of $(1)$. We have
\begin{align}
x & = t_1 x_1 + \cdots + t_k x_k \\[5pt]
y & = s_1 y_1 + \cdots + s_k y_k
\end{align}
for some scalars $t_1,\ldots,t_k,s_1,\ldots,s_k$, all $\ge0$ and satisfing $t_1+\cdots+t_k=1=s_1+\cdots+s_k$.
Let
$$
w_i = \frac{ct_i}{ct_i + (1-c)s_i} x_i + \frac{(1-c)s_i }{ct_i + (1-c)s_i} y_i\qquad \text{for }i=1,\ldots,k
$$
and
$$
r_i = c t_i + (1-c) s_i \qquad \text{for }i=1,\ldots,k.
$$
Then
$$
cx+(1-c)y = \sum_{i=1}^k r_i w_i. \tag 2
$$
Since it was assumed that $X_i$ is convex, we have $w_i\in X_i$ for $i=1,\ldots,k$. Therefore, $(2)$ is a member of the right side of $(1)$.
Best Answer
You can do the same thing with a half circle instead of the $x$-axis. Consider $D_\theta$ the closed segment joining $(0,0)$ to $(\cos\theta,\sin\theta)$, for $-\frac\pi2<\theta<\frac\pi2$. Then $$\bigcup\limits_{\theta\in(-\frac\pi2,\frac\pi2)} D_\theta=\{(x,y)\in\Bbb R^2\,:\, (x^2+y^2\le 1\land x>0)\lor x=y=0\}$$
More generally, consider your favourite bounded and not closed convex set $C$, and take $\{D_{x,y}\,:\, (x,y)\in C\times C\}$ where $D_{x,y}=\{tx+(1-t)y\,:\, t\in[0,1]\}$ is the closed segment joining $x$ to $y$. Then $$\bigcup_{(x,y)\in C\times C} D_{x,y}=C$$