TITLE: If $W_{1} \cup W_{2}$ is a subspace of $V$, then $W_{1} \subset W_{2}$ or $W_{2} \subset W_{1}$.
QUESTION [0 upvotes]: Is it possible to show that if $W_{1} \cup W_{2}$ is a subspace of $V$, then $W_{1} \subset W_{2}$ or $W_{2} \subset W_{1}$ using a direct proof? 
I've done the following:
Let $W_{1} \cup W_{2}$ be a subspace of $V$. Then $x,y \in W_{1} \cup W_{2}$ implies $(x+y) \in W_{1} \cup W_{2}$. Suppose $x \in W_{1}, y \in W_{2}$, and $(x+y) \in W_{2}$. Then since $W_{2}$ contains the inverse of $y$, the vector $-y$, we have $x = (x + y + (-y)) \in W_{2}$.

REPLY [2 votes]: You have to assume that $W_1$ and $W_2$ are subspaces of $V$. Otherwise this result is false.
 Assuming that $W_1$ and $W_2$ are subspaces of $V$ we can prove this result by contradiction. Suppose there exists $x \in W_1$ such that $x \notin W_2$  and $y \in W_2$ such that $y \notin W_1$. Then $x,y \in W_1\cup W_2$. Hence $x+y \in W_1\cup W_2$. If $x+y \in W_2$ then $x=(x+y)-y \in W_2$ (since $W_2$ is a subspace),  a contradiction. Similarly  $x+y \in W_1$ leads to a contradiction.
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