At the end of the game, the number of gold coins received by Bob less that of Alice is equal to $\sum_{i=1}^{n}(-1)^{|B_i \cap S|} \cdot a_i$. The number of Alice coins is greater than that of Bob if and only if $\sum_{i=1}^{n}(-1)^{|B_i \cap S|} \cdot a_i < 0$.

Suppose (for contradiction purposes) that for each nonempty set $S$, the number of Alice's coins is less that that of Bob, i.e.,
$$\sum_{i=1}^{n} (-1)^{|B_i \cap S|} \cdot a_i \geq 0.$$
Observe that if $S = \emptyset$, then $\sum_{i=1}^{n}(-1)^{|B_i \cap S|} \cdot a_i = \sum_{i=1}^{n} a_i > 0$. Adding over all possible sets $S$, we get
$$\sum_{S: S \subseteq \{1, 2, \dots, m\}} \sum_{i=1}^{n} (-1)^{|B_i \cap S|} \cdot a_i > 0. \quad (1)$$
Consider the sum $\sum_{S: S \subseteq \{1, 2, \dots, m\}} (-1)^{|B \cap S|}$ when $B$ is a nonempty subset of $\{1, \dots, m\}$. Writing
$$S = C \cup D \quad \text{where} \quad C = S \setminus B, D = S \cap B, C \cap D = \emptyset,$$
we have
$$\begin{aligned} \sum_{S: S \subseteq \{1, 2, \dots, m\}} (-1)^{|B \cap S|} &= \sum_{C: C \subseteq \{1, 2, \dots, m\} \setminus B} \sum_{D: D \subseteq B} (-1)^{|B \cap (C \cup D)|} \\ &= \sum_{C: C \subseteq \{1, 2, \dots, m\} \setminus B} \sum_{D: D \subseteq B} (-1)^{|D|}. \end{aligned}$$
As $|B| > 0$, we have $\sum_{D: D \subseteq B} (-1)^{|D|} = \sum_{r=0}^{|B|} (-1)^r \left(\begin{smallmatrix} |B| \\ r \end{smallmatrix}\right) = 0$. Thus, $\sum_{S: S \subseteq \{1, 2, \dots, m\}} (-1)^{|B \cap S|} = 0$ for each nonempty set $B \subseteq \{1, \dots, m\}$. Consequently,
$$\sum_{S: S \subseteq \{1, 2, \dots, m\}} \sum_{i=1}^{n} (-1)^{|B_i \cap S|} \cdot a_i = \sum_{i=1}^{n} \left( a_i \sum_{S: S \subseteq \{1, 2, \dots, m\}} (-1)^{|B_i \cap S|} \right) = 0,$$
which contradicts (1).