Given $2^{n-1}$ subsets of a set with $n$ elements with the property that any three have nonempty intersection, prove that ....

Given a family of $2^{n-1}$ subsets of a set $S=\{1,2,3,...,n-1,n\}$ with the property that any three have nonempty intersection. Prove that the intersection of all the sets in this family is nonempty.

I'm interested if the finishing of the solution here can be done like this:

As there was constructed $W$ is $n-1$ dimensional set in $\mathbb{F}_2^n$. Thus we have $W^{\bot}\oplus W =\mathbb{F}_2^n$ and $\dim W^{\bot} =1$, so $W^{\bot} = \{0,v\}$ for some vector $v\ne 0$. We can assume WLOG $$v = (\underbrace{1,1,1,...,1}_k,0,0,...,0)$$ Now each vector can be uniquely expressed with $v$ and some $w\in W$, in particulary we have $$e_1=(1,0,0,...0) =av+w$$ for some $w\in W$ and $a \in \mathbb{F}_2$. So if $a=0$ then $e_1\in W$ so $0=v\cdot e_1 =1$ which is impossible. Thus $a=1$ then $e_1-v \in W$ so $0=v\cdot (e_1-v) =1-k$ which means $k$ is odd.

So each vector not in $W$ is generated with odd number of vectors in $\{e_1,e_2,...e_k\}$ and an arbitrary number of vectors in $\{e_{k+1},e_{k+2},...e_n\}$ where $e_1,e_2,...,e_n$ represents the standard basis of $\mathbb{F}_2^n$.

Since the number of vectors not in $W$ is $2^{k-1} \cdot 2^{n-k} = 2^{n-1}$, we see if $k\geq 3$ then we have in $e_1$ and $e_2$ in $W^C$ so sets $\{1\}$ and $\{2\}$ are in $\mathcal{F}$ which is impossible. So $k=1$ and all sets in $\mathcal{F}$ have $1$ in common.

Non linear algebra solution: Prove that the intersection of all the sets is nonempty.

The first step $W^{\perp}\oplus W =\mathbb F_2^n$ is already problematic. For example, take $n=2$, and $W = \{(0,0), (1,1)\}$ and the "inner product" given by $\langle (a_1,a_2), (b_1, b_2) \rangle = a_1a_2 + b_1b_2$, then $W^{\perp} = W$, so $W+W^{\perp}=W$ which is not even a direct sum.

In general, we should not expect a non-degenerate bilinear form over $F^n$ behaves like an inner product over $\mathbb R$. Even over $\mathbb R$, only those that satisfy the positivity condition serve as inner product, but positivity cannot be defined over arbitrary fields.

In this particular case, we know that $W$ is essentially all vectors whose $i$-th component is $0$, so $W^{\perp} = e_i$, and $W\oplus W^{\perp}=\mathbb F_2^n$ holds, but this is unclear before we know the conclusion of the problem.