Suppose $A_1, A_2$ are affine subsets of $V$, then $A_1\cap A_2$ is either an affine subset of $V$ or $\emptyset$. The Problem: Suppose $A_1, A_2$ are affine subsets of $V$, then $A_1\cap A_2$ is either an affine subset of $V$ or $\emptyset$.
The Source: Sheldon Axler's Linear Algebra Done Right 3rd edition.
My Background: Sheldon Axler's Linear Algebra Done Right 3rd edition Section 1.A-Section 3.E.
My Attempt: Suppose $A_1=\vec{v}+V_1$, $A_2=\vec{w}+V_2$, where $V_1, V_2$ are subspaces of $V$.After trying a few options, I conjectured that if $A_1\cap A_2\neq\emptyset$, then $A_1\cap A_2=\frac{\vec{v}+\vec{w}}{2}+(V_1+V_2)$, where $V_1+V_2$ is the sum of $V_1$ and $V_2$.
Proving $A_1\cap A_2\subseteq\frac{\vec{v}+\vec{w}}{2}+(V_1+V_2)$ turns out to be quite straightforward; but I failed to show that $\frac{\vec{v}+\vec{w}}{2}+(V_1+V_2)\subseteq A_1\cap A_2$. Is my conjecture false? Any help would be greatly appreciated.

It is simpler than this! You don't have to consider subspaces! There are two cases
i)$A_{1}\bigcap\,A_{2}\neq\varnothing $ or
ii)$\,\,A_{1}\bigcap\,A_{2}=\varnothing$.
In case i) take $v,w \in A_{1}\bigcap\,A_{2}$. Then $v,w \in A_{1}$ which gives $\lambda v+(1-\lambda)w \in A_{1}$ where $\lambda \in \mathbb{R}$.But $v,w \in A_{2}$, hence $\lambda\,v+(1-\lambda)w\,\in\,A_{2}$.
Therefore for any $ v,w \in A_{1}\bigcap\,A_{2}$ we get $\lambda\,v+(1-\lambda)w\, \in A_{1}\bigcap\,A_{2}$ for any $\lambda \in \mathbb{R}$
which is the definition of an affine space! This is all we need!
