Jänich linear algebra, Question 2.3 solution clarification Let $ U_{1} $ and $ U_{2}$ be vector subspaces of a vector space $ V $. 
Prove if $ U_{1} \cup  U_{2} = V $, then $ U_{1} = V $ or $ U_{2} = V $ or both.
Attempt:  
$ U_{1} \cup U_{2} = V $ $\implies$ $ (U_{1} \cup U_{2} ) \setminus U_{2} = V \setminus U_{2} $. This is equivalent to $ U_{1} = V \setminus U_{2} $. But this is a contradiction as this means that $ 0 \notin U_{1} $. $ \square $
 A: I don't think your proof works. As Huy's comment suggests, there's a  mistake in your simplification:
 $$(U_1\cup U_2)\setminus U_2 = (U_1\setminus U_2)\cup (U_2\setminus U_2) = U_1\setminus U_2,$$
which is not necessarily equal to $U_1$.

As a side note, I want to point out that if you're using a proof by contradiction, you have to use the negation of your conclusion ($U_1 \neq V$ and $U_2 \neq V$ in this case) somewhere.

Here's a proof.
A union of two spaces is a subspace if and only if one subspace is contained in the other. (For a proof of this, see this question.)
$U_1\cup U_2=V$ is a subspace  $\implies$ either $U_1 \subseteq U_2$ or $U_2 \subseteq U_1$.
Say, without loss of generality, that $U_1\subseteq U_2$.
But then $V=U_1\cup U_2=U_2$.
A: suppose neither $U_1$ nor $U_2$ is $V$. then $\exists u_1 \in U_1 \setminus U_2$ and $\exists u_2 \in U_2 \setminus U_1$
consider $v=u_1+u_2$
since $v \in V = U_1 \cup U_2$ we must have $v \in U_1$ or $v \in U_2$. 
suppose $v \in U_1$ then $u_2= v-u_1$ also lies in $U_1$, contradicting the hypothesis. similarly in the case $v \in U_2$
