Suppose $U$ and $W$ are distinct four-dimensional subspaces of a vector space $V$, where $\dim V = 6$. Find the possible dimension of $U \cap W$.
Since $U$ and $W$ are distinct, shouldn't $U \cap W = 0$?
I think you're mistaking the word "distinct" for the word "disjoint".
The word "disjoint" means $U\cap W=\varnothing$, i.e. there are no elements in common between $U$ and $W$. Actually, it is impossible for two subspaces of a vector space to be disjoint, since they always must both contain the $0$ element. Thus $0\in U\cap W$ for any two subspaces $U$ and $W$. However, if the intersection of $U$ and $W$ is only the zero element and nothing else, this may sometimes be expressed in symbols as $U\cap W=0$.
The word "distinct" just means $U\neq W$, i.e. they aren't identical, but they can still have elements in common.
By Grassmann formula for two subspaces $U$ and $W$ of a vector space $V$: $$\dim (U\cap W) = \dim U+\dim W-\dim(U+W)$$ Here, $\dim U=\dim W = 4$ and $4\leq \dim(U+W)\leq 6$ since $W\subseteq U+W\subseteq V$. Thus $-6\leq -\dim(U+W)\leq -4$ and we obtain $$2 = 8-6 \leq \dim (U\cap W) = 8-\dim(U+W) \leq 8-4 = 4.$$
Finally, $\dim(U\cap W)=4$ means that $U\cap W=U=W$ (for dimension reasons), hence $\dim(U\cap W) = 2$ or $3$. Thus, $U\cap V$ can't be $0$.
In fact, distinct subspaces are simply different (i.e. there's a vector belonging to only one of them), but they can have a non-zero intersection.