I think I figured it out. Indeed, $V(\mathcal{C}')\leq n$. To see this let $\widetilde{V}:=\{(x_1,t_1),\ldots,(x_n,t_n)\}\subset X\times\mathbb{R}.$ Suppose there exists an $i\in\{1,\ldots,n\}$ such that $t_i>1$. Then $(x_i,t_i)\notin I_C$ for all $C\in\mathcal{C}$, thus $\mathcal{C}'$ doesn't pick out $\{(x_i,t_i)\}$ and therefore does not shatter $\widetilde{V}$. If there exists an $i\in\{1,\ldots,n\}$ such that $t_i\leq 0$ then $(x_i,t_i)\in I_C$ for all $C\in\mathcal{C}$. But in this case setting $j=\{1,\ldots,n\}\backslash\{i\}$ results in $\{(x_j,t_j)\}\neq \widetilde{V}\cap I_C$ for all $C\in\mathcal{C}$. Therefore, $\mathcal{C}'$ doesn't pick out $\{(x_j,t_j)\}$ and consequently doesn't shatter $\widetilde{V}$. We will further assume that $t_i\in(0,1]$ for all $i\in\{1,\ldots,n\}$.
Since $V(\mathcal{C})=n$ there exists a subset $A$ of $V:=\{x_1,\ldots,x_n\}$, $A:=\{x_i:i\in T\subset\{1,\ldots,n\}\}$ such that $\mathcal{C}$ doesn't pick out $A$, i.e. $A\neq V\cap C$ for all $C\in\mathcal{C}$. Denote $\widetilde{A}:=\{(x_i,t_i):i\in T\subset\{1,\ldots,n\}\}$. Let $C\in\mathcal{C}$. Then $A\neq V\cap C$ and this means that:
- there exists $i\in\{1,\ldots,n\}$ such that $x_i\in A\backslash C$, or
- there exists $i\in\{1,\ldots,n\}$ such that $x_i\in C\backslash A$.
In the first case, from $t_i>0$ it follows that $(x_i,t_i)\notin I_C$, therefore $\widetilde{A}\neq \widetilde{V}\cap I_C$. In the second case $(x_i,t_i)\in I_C$, thus again $\widetilde{A}\neq \widetilde{V}\cap I_C$. Since $C$ is arbitrary $\widetilde{A}\neq \widetilde{V}\cap I_C$ for all $C\in\mathcal{C}$, i.e. $\mathcal{C}'$ doesn't pick out $\widetilde{A}$ and consequently doesn't shatter $\widetilde{V}$.
Suppose $V(\mathcal{C}')=n$. We will show that $V(\mathcal{C})\leq n$. Let $\{x_1,\ldots,x_n\}\subset X$. Denote $a:=1/2$. By $V(\mathcal{C}')=n$, the set $\{(x_1,a),\ldots,(x_n,a)\}$ has a subset $\{(x_i,a):i\in T\}$, $T\subset\{1,\ldots,n\}$, such that \begin{align*} \{(x_i,a):i\in T\}\neq \{(x_1,a),\ldots,(x_n,a)\}\cap I_C,\ \forall C\in\mathcal{C}. \end{align*} Let $C\in\mathcal{C}$. Then
- there exists $i\in T$ such that $(x_i,a)\in \{(x_i,a):i\in T\}\backslash I_C$, or
- there exists $i\in\{1,\ldots,n\}\backslash T$ such that $(x_i,a)\in I_C$.
In the first case $(x_i,a)\notin I_C$, which can only be true when $x_i\notin C$, therefore $\{x_i:i\in T\}\neq \{x_1,\ldots,x_n\}\cap C$. In the second case, from $(x_i,a)\in I_C$ it follows that $x_i\in C$, however, by the choice of $i$, $x_i\notin\{x_i:i\in T\}$, therefore $\{x_i:i\in T\}\neq \{x_1,\ldots,x_n\}\cap C$. Since $C$ is arbitrary, $\{x_i:i\in T\}\neq \{x_1,\ldots,x_n\}\cap C$ for all $C\in\mathcal{C}$, i.e. $\mathcal{C}$ doesn't pick out $\{x_i:i\in T\}$ and consequently doesn't shatter $\{x_1,\ldots,x_n\}$.