Show that $U\cup W$ is a subspace of $V$ if and only if $U \subset W$ or $W \subset U$ 
Let $U$ and $W$ be subspaces of $V$. Show that $U\cup W$ is a subspace of $V$ if  and only if $U \subset W$ or $W \subset U$.

I am not sure what I can do with the assumption that $U\subset W$ or $W \subset U$ to get started. If I could get a hint pointing me in the right direction, that would be appreciated, thanks
 A: Suppose that $U\cup W$ is a subspace. There are two possibilities


*

*If $U\subseteq W$ then we are done.

*If not, let $u\in U\setminus W$ be fixed. We will show that $W\subseteq U$. For every $w\in W$, we have both $u\in U\cup W$ and $w\in U\cup W$, so $u+w\in U\cup W$. Since $W$ is a subspace, $w\in W$ and $u\not\in W$, then $u+w\not\in W$. Therefore $u+w\in U$. Since $u\in U$, then $w=(u+w)-u\in U$ as well.
In both cases, $U\subseteq W$ or $W\subseteq U$.
A: Let us assume that $U\not\subset W$ and $W \not\subset U$ and $U\cup W$ is a subspace. 
Then there exist $v \in U$ such that $v\not\in W$ and $w\in W $ such that $w \not\in U$, since $U \cup W $ is a subspace $u+w \in U \cup W$. This implies that $u+w$ belongs to $U$ or in $W$.
But this would force $v \in W$ or $w \in U$. Hence the assertion follows.  
A: One direction is straightforward.
Suppose $U \cup W$ is a subspace and $W \not \subset U$. Then there is some $w_0 \in W \setminus U$.
Now pick $u \in U$ and note that $u-w_0 \in U \cup W$ (since a subspace) and
$u-w_0 \in W $ (otherwise we would have $w_0 \in U$). Hence $u = (u-w_0) + w_0 \in W$
and so $U \subset W$.
A: Proof $U\setminus W\neq\{\emptyset\}$ and $W\setminus U\neq\{\emptyset\}$ implies $U\cup W$ not a subspace of $V$
Suppose there are $u\in U\setminus W$ and $w\in W\setminus U$. Then $u+w\notin U\cup W$ (we know $u+w\notin U$ because that would imply $w\in U$, being $U$ a subspace, and we can show similarly that $u+w\notin W$).
Therefore, there are $u,w\in U\cup W$ such that $u+w\notin U\cup W$, hence $U\cup W$ is not closed under addition and thus not a subspace.
Proof that $U\cup W\le V$ implies $U\subseteq W$ or $W\subseteq U$
This is just another way to prove the same statement as above.
For all $u\in U, w\in W$, we have $u+w\in U\cup W$. For each such $u,w$, either $u+w\in U$ or $u+w\in W$ (or both). If $w\notin U$, then $u+w\notin U$ (because $u+w\in U$ would imply $(u+w)-u=w\in U$), and therefore $u+w\in W$, which implies $u\in W$.
We conclude that the existence of some $w\in W\setminus U$ implies $u\in W$ for all $u$, that is, $U\subseteq W$. Similarly, one proves that if $U\subseteq W$ is false then $W\subseteq U$.
