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?

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 $$N \subseteq \bigcap_{i=1}^n N_i = \text{span}(B \setminus \{v_1,\ldots,v_n\}) = \text{span}(\emptyset) = \{0\}$$ 
