direct proof of $U_1\cup U_2$ is subspace of $V$ $\implies$ $U_1\subset U_2$ or $U_2\subset U_1$ I'm trying to prove the following by direct proof (I've already got a proof by contradiction).


  The union of two subspace of $V$ is a subspace of $V$ implies one of the subspaces is contained in the other.


Here's how far I've gotten:  Suppose $U_1\cup U_2$ is a subspace of $V$. Let $u_1 \in U_1$ and $u_2 \in U_2$.  Then, by closure of vector addition in $U_1\cup U_2$, $u_1 + u_2 \in U_1\cup U_2$.  This implies that $u_1 + u_2 \in U_1$ or $u_1 + u_2 \in U_2$.  WLOG, suppose $u_1 + u_2 \in U_1$.  Then $u_2 = (u_1 + u_2) - u_1 \in U_1$.

At this point, I'd like to be able to say that because $u_2$ was chosen arbitrarily, this implies that $U_2 \subset U_1$.  But I also made the choice that $u_1 + u_2 \in U_1$, so for some other choice of $u_2$ it could be the case that $u_1 + u_2 \in U_2$ which would imply $u_1 \in U_2$.

Is there any way to complete this proof directly or do I have to stick with the proof by contradiction for this one?
You don't need to use contradiction really, but mere inspection of cases. Just prove that if $U_1$, say, is not contained in $U_2$ i.e. if there's a $u_1\in U_1\smallsetminus U_2$, then $U_2\subset U_1$.
The idea is OK, go by contradiction.

Suppose  neither inclusion holds, but $U_1 \cup U_2$ is a subspace.
So we have $u_1 \in U_1 ,u_1 \notin U_2$ and $u_2 \in U_2, u_2 \notin U_1$ to witness the non-inclusions. 

But $u_1 + u_2 \in U_1 \cup U_2$, say in $(u_1 + u_2) \in U_i$.
But then $(u_1 + u_2 ) - u_i \in U_i$, while it equals $u_j$ where $j \neq i$. Contradiction, as $u_j \notin U_i$ when $j \neq i$.