I posted this before but there was some confusion regarding the wordings. The following question is from Linear Algebra, by Bhimasankaram & Rao (2000).
Show that the union of two flats is a flat iff one of them is contained in the other or $|F| = 2$ and one is a translate of the other.
The following is a definition of a flat.
$\textbf{Defintion}~$ A set $A \subseteq V$ is said to be a flat or an affine set iff either $A = \emptyset$ or $A = \vec x + S$ for some subspace $S$ of $V$ and some $\vec x \in V$.
I use the following theorem in my incomplete proof:
$\textbf{Theorem}~$ $A \subseteq V$ is a flat iff $A$ is closed under affine combinations.
I have already proved the sufficiency, so assume it has already been proven. Here is my following incomplete proof for the necessity:
Let $A,B$ be flats in a vector space $V$ over $F$, and assume that $A \cup B$ is also a flat. Then if $A \subseteq B \lor B \subseteq A$, certainly $A \cup B$ is a flat. So, let them not be subsets.
Also assume that $|F| > 2$. Now, my only exposure to fields is this LA book, so I may be wrong, but there must exist some $c$ in $F$ such that it is a non-zero element that is not an additive inverse of $1$. My reasoning is that if the third element is not an additive inverse of $1$, then it is $c$, but if it is, then $1$ is a non-zero element that is not an additive inverse of $1$. Also, $c \neq 0 \implies -c \neq 0$. So, consider the affine combination of $\vec a \in A - B$ and $\vec b \in B - A$: $$ \vec v = (1+c) \vec a + (-c) \vec b \in \operatorname{aff} (A \cup B) = A \cup B \tag{$\because$ $A \cup B$ is a flat} $$
$\vec v \in A \cup B \iff \vec v \in A \lor \vec v \in B$, so if $\vec v \in A$ consider the following affine combination: \begin{align} \left(\frac{-1}c\right) \vec v + \left(\frac{1+c}c\right) \vec a = \vec b \end{align} Therefore $\vec b$ is an affine combination from $A$ and must be in $A$ which is a contradiction. Similarly, we can find a contradiction $\vec v \in B \implies \vec a \in B$, as follows: \begin{align} \left(\frac{1}{1+c}\right) \vec v + \left(\frac{c}{1+c}\right) \vec b = \vec a \end{align}
Therefore $A \cup B$ must not contain all affine combination and is hence, not a flat.
Now what happens if $|F|=2$? Certainly in $\{0,1\}^2$, $\{ (0,0), (1,0) \}$ and $\{ (0,1), (1,1) \}$ are flats whose union is also a flat, and certainly one is a translate of the other, but how do I prove this is necessary?
Suppose neither of $A$ and $B$ is contained in the other. After a translation, we may assume that $0\in A\setminus B$, so $A$ is actually a linear subspace of $V$ and so is $A\cup B$. Now fix an element $x\in B\setminus A$; I claim that $B$ is equal to the translate $x+A$. For any $y\in A$, $x+y\in A\cup B$ since $A\cup B$ is a linear subspace of $V$. But $x+y$ cannot be in $A$, since otherwise $(x+y)-y=x$ would be in $A$, so $x+y\in B$. That is, $x+A\subseteq B$.
Conversely, suppose $y\in B$; we wish to show that $y-x\in A$ so that $y\in x+A$. First, since $A\cup B$ is a linear subspace, we know $y-x\in A\cup B$. If $y-x$ is in $B$, then all of $y-x,y,$ and $x$ are in $B$, so the affine combination $1\cdot(y-x)+1\cdot y+ (-1)\cdot x=0$ is in $B$ as well. This contradicts our assumption that $0\in A\setminus B$. Thus $y-x$ cannot be in $B$, so it must be in $A$, as desired.