VC dimension of hypothesis space of finite union of intervals
VC dimension is defined for a hypothesis space $H$, e.g. a set of binary classifiers $C \rightarrow \{0, 1\}$. For example, hypothesis space
$$H=\{{\Bbb 1}_{x \le \theta}: \theta \in {\Bbb R}\}$$
has VC dimension $1$, because for any $C=\{a&lt;b\}$,  it does not contain a classifier that gives $\{a \rightarrow 0, b\rightarrow 1\}$. 

For example, a classifier from $H$ would be $f(x)={\Bbb 1}_{x \le a}$ that gives  $\{a \rightarrow 1, b \rightarrow 0\}$.

From C to H

As you have illustrated in the comments, we can build a hypothesis space $H$ from $C$ as follows:
$$H=\left\{{\Bbb 1}_{x \in C}: C = \left\{\bigcup_{i=1}^{k}(a_i, b_i): a_i, b_i \in {\Bbb R}, a_i &lt; b_i, i=1,2,..,k\right\}\right\}$$
Meaning, each classifier in $H$ is a union of $k$ intervals that labels a point inside the union as $1$ and outside as $0$.

VC dimension of this $H$ is $2k$:


For VC $\geq 2k$: Let $A$ be an arbitrary set , and $A \rightarrow \{0, 1\}$ be an arbitrary labeling. By going from minimum to maximum member of $A$, we can cover all adjacent $1$s with one interval, and only need to use another interval when there is a $0$ barrier. Therefore, we need $k$ intervals to cover $k$ isolated regions of $1$s. Furthermore, a set with  $2k$ members has at most $k$ isolated $1$s (since to have $k+1$ isolated $1$s there should be $k$ $0$ barriers in-between), and thus, needs at most $k$ intervals.
For VC $&lt; 2k+1$ by contradiction: for any ordered set $A_{2k+1}=\{a_1&lt;...&lt;a_{2k+1}\}$, there is labeling $a_k \rightarrow 1_{\text{k odd}}$, i.e.  $\{a_1 \rightarrow 1, a_2 \rightarrow 0,...,a_{2k+1} \rightarrow 1\}$ with  $k+1$ isolated $1$s which cannot be covered with $k$ intervals.