Let $\set{\vect{u}_1,\,\vect{u}_2,\,\vect{u}_3}$ and $\set{\vect{v}_1,\,\vect{v}_2,\,\vect{v}_3}$ be bases for $U$ and $V$ (respectively). Then, the set $\set{\vect{u}_1,\,\vect{u}_2,\,\vect{u}_3,\,\vect{v}_1,\,\vect{v}_2,\,\vect{v}_3}$ is linearly dependent, since Theorem G says we cannot have 6 linearly independent vectors in a vector space of dimension 5. So we can assert that there is a nontrivial relation of linear dependence,
\begin{equation*}
a_1\vect{u}_1+a_2\vect{u}_2+a_3\vect{u}_3+b_1\vect{v}_1+b_2\vect{v}_2+b_3\vect{v}_3=\zerovector
\end{equation*}
where $a_1,\,a_2,\,a_3$ and $b_1,\,b_2,\,b_3$ are not all zero.
We can rearrange this equation as
\begin{equation*}
a_1\vect{u}_1+a_2\vect{u}_2+a_3\vect{u}_3=-b_1\vect{v}_1-b_2\vect{v}_2-b_3\vect{v}_3
\end{equation*}
This is an equality of two vectors, so we can give this common vector a name, say $\vect{w}$,
\begin{equation*}
\vect{w}=a_1\vect{u}_1+a_2\vect{u}_2+a_3\vect{u}_3=-b_1\vect{v}_1-b_2\vect{v}_2-b_3\vect{v}_3
\end{equation*}
This is the desired nonzero vector, as we will now show.
First, since $\vect{w}=a_1\vect{u}_1+a_2\vect{u}_2+a_3\vect{u}_3$, we can see that $\vect{w}\in U$. Similarly, $\vect{w}=-b_1\vect{v}_1-b_2\vect{v}_2-b_3\vect{v}_3$, so $\vect{w}\in V$. This establishes that $\vect{w}\in U\cap V$ (Definition SI).
Is $\vect{w}\neq\zerovector$? Suppose not, in other words, suppose $\vect{w}=\zerovector$. Then
\begin{equation*}
\zerovector=\vect{w}=a_1\vect{u}_1+a_2\vect{u}_2+a_3\vect{u}_3
\end{equation*}
Because $\set{\vect{u}_1,\,\vect{u}_2,\,\vect{u}_3}$ is a basis for $U$, it is a linearly independent set and the relation of linear dependence above means we must conclude that $a_1=a_2=a_3=0$. By a similar process, we would conclude that $b_1=b_2=b_3=0$. But this is a contradiction since $a_1,\,a_2,\,a_3,\,b_1,\,b_2,\,b_3$ were chosen so that some were nonzero. So $\vect{w}\neq\zerovector$.
How does this generalize? All we really needed was the original relation of linear dependence that resulted because we had “too many” vectors in $W$. A more general statement would be: Suppose that $W$ is a vector space with dimension $n$, $U$ is a subspace of dimension $p$ and $V$ is a subspace of dimension $q$. If $p+q>n$, then $U\cap V$ contains a nonzero vector.