Proof. Let $e_S$ denote the characteristic vector of $S$, whose coefficients $e_i$ are 1 if $i\in S$ and 0 otherwise. Consider the following primal linear program on the left and its dual on the right.
\begin{equation*} \begin{aligned} v_P(S)&=\min e_\nn\cdot u\\[2pt] \text{s.t.}\qquad&\begin{aligned}[c] e_S\cdot u&\geq v(S)\text{ for all $S\subsetneq\nn$,}\\[2pt] e_\nn\cdot u&\geq v(\nn) \end{aligned} \end{aligned} \end{equation*}
\begin{equation*} \begin{aligned} v_P(S)&=\max\sum_S\lambda_Sv(S)\\ \text{s.t.}\qquad&\begin{aligned}[t] \sum_{S\ni i}\lambda_S+\eta&=1\text{ for all $i\in\nn$,}\\ \lambda,\eta&\geq 0 \end{aligned} \end{aligned} \end{equation*}
The set of feasible primal solutions is bounded below the the $v(\{i\})$, so the primal problem, and therefore the dual, have optimal solutions.
Suppose the core is empty. Then it must be true of any optimal solution $u^*$ that $e_\nn\cdot u^*>v(\nn)$. If $e_\nn\cdot u^*>v(\nn)$, then it must be true that for all feasible solutions $u$, $e_\nn\cdot u>v(\nn)$, and so the core is empty.