The problem is:
Let U and V be two subspaces of $R^n$. Show that if $$\dim(U) + \dim(V) > n$$ then there must be a non-zero vector in their intersection, i.e. $U \cap V \neq \{0\}$
So I know that if I take the example of $R^2$, I can let $U$ be a line in $R^2$ and $V$ be a subspace that spans all of $R^2$. It's clear that the line U is the intersection...
But how do I formally mathematically prove this in the $R^n$ case?
As the comments say, this is a direct result of the theorem relating the dimensions of the direct sum and intersection of two subspaces. There are lots of different ways to talk about linear algebra, so here's a proof by contraposition.
Assume $U\cap V=\{0\}$. Choose $B$ to be a basis of $U$ and $B'$ to be a basis of $V$. Since $U\cap V=\{0\}$, $B\cup B'$ is a linearly independent set of $\dim U+\dim V$ vectors. But those are all vectors in $\mathbb R^n$, which obviously has dimension $n$. Therefore, $\dim U+\dim V\leq n$.