Not sure if this is the easiest way to do things, but I'll try a contradiction. Assume for the purposes of contradiction $\displaystyle \bigcap_{W \in \mathcal{A}} W = \{0 \}.$

First, there are no trivial subspaces $\displaystyle W$ in $\displaystyle \mathcal{A}$, as if there were, we would take $\displaystyle \mathcal{B} = \{W\} \subseteq \mathcal{A}$ with $\displaystyle |\mathcal{B}| = 1$ and we have $\displaystyle \bigcap_{W \in \mathcal{B}} W = \{0 \},$ a contradiction.

There must also exist at least two elements in $\displaystyle \mathcal{A}$ because if there were only one, we could again take $\displaystyle \mathcal{B} = \mathcal{A}$ with $\displaystyle |\mathcal{B}| = 1$ and, from our assumption about $\displaystyle \mathcal{A}$, we have $\displaystyle \bigcap_{W \in \mathcal{B}} W = \{0 \}$, a contradiction.

Now, if we take any $\displaystyle \alpha \in W_i$, there must be a $\displaystyle W_j$ such that $\displaystyle \alpha \not\in W_j$ for a $\displaystyle i \ne j$ because $\displaystyle \bigcap_{W \in \mathcal{A}} W = \{0 \}$ by assumption. Thus $\displaystyle \bigcap W_i + W_j = \{0 \}$. But since we can take $\displaystyle \mathcal{B} = W_i + W_j$ and $\displaystyle |\mathcal{B}| = 2 \le n$, we have $\displaystyle \bigcap_{W \in \mathcal{B}} W = \{0 \}$, a contradiction.

Therefore, $\displaystyle \bigcap_{W \in \mathcal{A}} W \ne \{0 \}.$