Subspace proof, just need a little push

Let $\displaystyle W_1 , W_2$ be non-empty subspaces of V. Prove that if $\displaystyle W_1 \cup W_2$ is a subspace of V, than either (I don't know how to do this in Latex) $\displaystyle W_1 \le W_2$ or $\displaystyle W_2 \le W_1$. ( $\displaystyle \le$ symbol should mean proper subset here).

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This should be done by contradiction I think. So assume that there exists $\displaystyle x_1 \in W_1$ and $\displaystyle x_2 \in W_2$, wuch that $\displaystyle x_1 , x_2 \notin W_1 \cap W_2$.

Still assume now that $\displaystyle W_1 \cup W_2$ is a subspace of V.

Now I'm having a brain fade. Hint perhaps?