Consider a special matrix $D$, $D$ has $m$ columns (one column for each set) and $n$ rows (for every number from $[n]$ ) such that-
$D_{ij} = \left\{\begin{matrix} 1 & \text{if } i \in A_j & & \\ 0& \text{otherwise}& & \end{matrix}\right.$ i.e $D_{ij} $ indicates presence/absence of $i^\text{th}$ element in $j^\text{th}$ set.
Now consider $D^TD$ and convince yourself that diagonal entries of $D^TD$ represents $|A_i |$ and non diagonal entries represents $|A_i \cap A_j|$.
$D^TD = \begin{pmatrix} .& & & & & & |A_i \cap A_j| \\ & .& & & & & \\ & & .& & & & \\ & & & |A_i|& & & \\ & & & &. & &\\ & & & & &.& \\ & & & & & & \end{pmatrix} = \begin{pmatrix} .& & & & & & \text{even}\\ & .& & & & & \\ & & .& & & & \\ & & & \text{odd}& & & \\ & & & &. & &\\ & & & & &.& \\ & & & & & & \end{pmatrix}$
Notice that: $\operatorname{rank}(D^TD)\leq \operatorname{rank}(D)$ $ \le n$
$\Big ( \because \small {\operatorname{rank}(AB)\leq min(\operatorname{rank}(A),\operatorname{rank}(B)) \text{ and } \operatorname{rank} A_{m\text{x}n} \le\min(m,n) }\Big) $
Claim: $\operatorname{rank}(D^TD)= m \text{ i.e } { det(D^TD) \neq 0}$
This claim follows $m \le n$.
proof - By design we see that $D^TD \equiv I \text{ mod }2$,
So, $\operatorname {det}(D^TD) \equiv \operatorname {det}(I) =1 \text{ mod }2 $
$\Rightarrow \operatorname {det}(D^TD) \neq 0$