(Note: edited after the first comment)
Let $A, B$ be nonempty subspaces of an affine space $E$ such that $A \cap B = \emptyset$.
I am asked to prove that there are $a \in A$ and $b \in B$ such that $\vec{ab}$ is orthogonal to both $A$ and $B$.
But take $A, B$ such that $A \cup B = E$. Then $A^\bot=B$ and $B^\bot=A$, so that $A^\bot \cap B^\bot = \emptyset$, and there's no vector with the required property. Isn't that a counterexample?
That case aside (!), for simple examples I can see how to choose two points so that the vector is orthogonal to both subspaces, but I don't know how to prove it in general.
I'd appreciate any hint of where I'm wrong or how the proof should go. Thanks in advance!
Hint.
Name $\vec{A},\vec{B}$ the underlying vector spaces of $A,B$. As $A \cap B =\emptyset$, $\vec{A} +\vec{B}$ is a proper linear subspace of $\vec{E}$, take $\vec{V} =(\vec{A} +\vec{B})^\perp$. $(A+\vec{V}) \cap B$ is not empty. Take $b \in (A+\vec{V}) \cap B$ and $a \in (b+\vec{V}) \cap A$.
$\vec{ab}$ is orthogonal to both $A$ and $B$.