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 $\{0\}$. 
$$$$ 
I have done the following: 
We suppose that there is a non-zero vector, say $u$, common to all $(n-1)$-dimensional subspaces. 
Then suppose that $B$ is the basis of the intersection, then $u\in B$. 
We can exchange $u$ by an other element $v\in V$ and we get again a basis, right? 
How can we continue?
 A: I am not sure this is the easiest way to answer your original question. The way to do what you’re asking is:
Asume by contradiction that the intersection has a non zero vector $u$. Complete $u$ to a basis $u, v_2, ....v_n$. Now $v_2, ...,v_n$ spans a $n-1$ subspace. $u$ can’t be an element of that sub space because it’s independent of $v_2, ...,v_n$ by our construction. This leads to the desired contradiction.
A: Let call N "the intersection of all subspaces of $V$ with dimension $n-1$" and $$N_i = \text{span}(B \setminus \{v_i\})$$
Then each $N_i$ is a subspace of $V$ with dimension $n-1$ and ofc it holds $$N \subseteq \bigcap_{i=1}^n N_i$$
Now it's pretty clear (or can be seen by induction) that $$N_i \cap N_j = \text{span}(B \setminus \{v_i,v_j\})$$
Hence $$N \subseteq \bigcap_{i=1}^n N_i = \text{span}(B \setminus \{v_1,\ldots,v_n\}) = \text{span}(\emptyset) = \{0\}$$ 
A: Since $B$ is a basis, you have
$$
\bar{v}=\sum_{i=1}^n \alpha_iv_i
$$
and, since $\bar{v}\ne0$, one of the coefficients $\alpha_i$ must be nonzero. Let it be $\alpha_j$. Then
$$
v_j=\frac{1}{\alpha_j}\biggl(\bar{v}-\sum_{i\ne j}\alpha_iv_i\biggr)
$$
implying that $v_j$ belongs to the span of $B'=\{\bar{v},v_1,\dots,v_{j-1},v_{j+1},v_n\}$. Clearly all $v_i$, with $i\ne j$, belong to the span of $B'$, so the span of $B'$ is the same as the span of $B$, that is, $V$. Hence $B'$ is a basis, because it is a spanning set with $n$ elements.
Since $\bar{v}$ does not belong to the span of $\{v_1,\dots,v_{j-1},v_{j+1},v_n\}$ which has dimension $n-1$, you conclude $\bar{v}$ does not belong to every $(n-1)$-dimensional subspace.

An easier proof.
Let $u$ be a nonzero vector in $V$. Complete it to a basis $\{u_1=u,u_2,\dots,u_n\}$.
Define $f\colon V\to\mathbb{R}$ by $f(u)=1$ and $f(u_i)=0$ for $i=2,\dots,n$.
Then $\ker f$ has dimension $n-1$ and $u\notin\ker f$.
