The intersection of all subspace of $V$ is $\{0\}$. Let $V$ be a $\mathbb{R}$-subspace with basis $B=\{v_1 ,v_2, \ldots , v_n\}$ and $\overline{v}\in V$, $\overline{v}\neq 0$. 
I have shown that if we exchange $\overline{v}$ with a $v_i\in B$ we get again a basis. 
I want to show, using this fact, that the intersection of all subspace of $V$ of dimension $n-1$ is $\{0\}$. 
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I have done the following: 
We suppose that there is a non-zero vector, say $u$, common to all $(n-1)$-dimensional subspaces. 
Then suppose that $B$ is the basis of the intersection, then $u\in B$. 
We can exchange $u$ by an other element $v\in V$ and we get again a basis, right? 
How can we continue?

I am not sure this is the easiest way to answer your original question. The way to do what you’re asking is:
Asume by contradiction that the intersection has a non zero vector $u$. Complete $u$ to a basis $u, v_2, ....v_n$. Now $v_2, ...,v_n$ spans a $n-1$ subspace. $u$ can’t be an element of that sub space because it’s independent of $v_2, ...,v_n$ by our construction. This leads to the desired contradiction.
