Positivity of the alternating sum associated to at most five subspaces

Here is a purely combinatorial proof for $n=5$. We first generalize the problem as follows:
Let $B_5$ be the boolean lattice of rank $5$, i.e. the subsets lattice of $\{1,2,3,4,5\}$.

Lemma: Let $\phi: B_5 \to \mathbb{R}_{\ge 0}$ be a map satisfying that $\forall a, b \in B_5$:

$(1)$ $ \ $ $a \le b \Rightarrow \phi(a) \le \phi(b)$ [poset morphism]
$(2)$ $ \ $ $\phi(a \vee b) + \phi(a \wedge b) \ge \phi(a ) + \phi(b)$

and let $a_i= \{i \}^\complement$ the complement of $\{i \}$ in $\{1,2,3,4,5\}$, then $$\sum_{r=1}^5 (-1)^{r+1}\sum_{i_1 < i_2 < \cdots < i_r} \phi( a_{i_1} \wedge \cdots \wedge a_{i_r}) \ge 0$$ proof: we reorganize the alternative sum into the sum of the following components:

• $\phi(\{1,2,3,4\}) - \phi(\{1,2,3\}) - \phi(\{1,2,4\}) + \phi(\{1,2\})$
• $\phi(\{1,3,4,5 \}) - \phi(\{1,3,4\}) - \phi(\{1,3,5\}) + \phi(\{1,3\})$
• $\phi(\{2,3,4,5\}) - \phi(\{2,3,4\}) - \phi(\{3,4,5\}) + \phi(\{3,4\})$
• $\phi(\{1,2,4,5 \}) - \phi(\{2,4,5\}) - \phi(\{1,4,5\}) + \phi(\{4,5\})$
• $\phi(\{1,2,3,5\}) - \phi(\{1,2,5\}) - \phi(\{2,3,5\}) + \phi(\{2,5\})$
• $ \phi(\{1,5\}) - \phi(\{1\})$
• $ \phi(\{2,4\}) - \phi(\{2\})$
• $ \phi(\{2,3\}) - \phi(\{3\})$
• $ \phi(\{1,4\}) - \phi(\{4\})$
• $ \phi(\{3,5\}) - \phi(\{5\})$
• $ \phi(\emptyset) $

but the first five components are positive by $(2)$, the next five components are positive by $(1)$, and the last is positive by definition $\square$.

Now the answer of the question is yes by observing that the map $\phi$ defined by $$\phi( a_{i_1} \wedge \cdots \wedge a_{i_r}) = \dim (V_{i_1} \cap \cdots \cap V_{i_r})$$ checks $(1)$ and $(2)$. For $(1)$ it is immediate. For $(2)$ we use the following equality and inclusion: $\dim(U+V) = \dim(U) + \dim(V) - \dim(U \cap V)$ and $(A\cap B ) + (A\cap C) \subseteq A$.