Vector spaces and intersections I was thinking of the following problem lately:

Suppose $V_1,V_2,V_3,V_4$ are vector subspaces of $\Bbb{R}^4$ of dimension $2$ such that $V_i\cap V_j=\{0\}$ for $i \neq j$. Is it true that we can find a two dimensional vector subspace of $\Bbb{R}^4$ such that $\dim V_i\cap W =1$ for $i=1,2,3,4$?

The same problem with all dimensions doubled was given to a Miklos Schweitzer competition in 2012.
Using a linear automorphism of $\Bbb{R}^4$ we can assume that $V_1=span\{e_1,e_2\},V_2=span\{e_3,e_4\}$ where $(e_1,e_2,e_3,e_4)$ is the canonical base. It is rather easy to construct $W$ which satisfies $\dim W\cap V_i = 1$ for $i=1,2,3$ but I didn't manage connecting it to the fourth space.
 A: $W$ may not always exist. Here is a counter-example. 
Following your notations, let $\{e_1,e_2,e_3,e_4\}$ be a basis of $\Bbb  R^4$ and let
$$V_1={\rm span}\{e_1, e_2\}, \quad V_2={\rm span}\{e_3, e_4\}.$$
Moreover, let
$$V_3={\rm span}\{e_1+ e_3, e_2+e_4\},\quad V_4={\rm span}\{e_1-e_4, e_2+e_3\}.$$ 
It is easy to check that those subspaces satisfy all the requirements. Now suppose $W$ is a two dimensional subspace of  $\Bbb  R^4$ and $\dim (W\cap V_1 )=\dim (W\cap V_2)=1$, so there are $a,b,c,d\in \Bbb R$, such that
$$ W={\rm span}\{ a e_1+be_2, c e_3+de_4\}.$$
Consider $u:=(a,b)$ and $v:=(c,d)$ as non-zero vectors in $\Bbb R^2$. Then direct calculation shows that $\dim (W\cap V_3)=1$ if and only if $u$ and $v$ are parallel, and $\dim (W\cap V_4)=1$ if and only if  $u$ and $v$ are perpendicular, i.e. those two conditions cannot be both satisfied simultaneously. 
A: What about
$$
V_1=span\{e_1,e_2\}, \quad V_2=span\{e_3,e_4\}, 
$$
$$
V_3=span\{e_1+e_3,e_2+e_4\}, \quad V_4=span\{e_1-e_3,e_2-e_4\}.
$$
These are all two-dimensional subspaces with $V_i\cap V_j=\{0\}$ for $i\ne j$.
Taking
$$
W=span\{e_1+e_2,e_3+e_4\}
$$
should answer the question: $dim(W\cap V_i)=1$ for all $i$.
