Union of two vector subspaces not a subspace?

The reason why this can happen is that all vector spaces, and hence subspaces too, must be closed under addition (and scalar multiplication). The union of two subspaces takes all the elements already in those spaces, and nothing more. In the union of subspaces $W_1$ and $W_2$, there are new combinations of vectors we can add together that we couldn't before, like $v_1 + w_2$ where $v_1 \in W_1$ and $w_2 \in W_2$.

For example, take $W_1$ to be the $x$-axis and $W_2$ the $y$-axis, both subspaces of $\mathbb{R}^2$.
Their union includes both $(3,0)$ and $(0,5)$, whose sum, $(3,5)$, is not in the union. Hence, the union is not a vector space.

The union of two subspaces is a subspace if and only if one of the subspaces is contained in the other.

The "if" part should be clear: if one of the subspaces is contained in the other, then their union is just the one doing the containing, so it's a subspace.

Now suppose neither subspace is contained in the other subspace. Then there are vectors $x$ and $y$ such that $x$ is in the first subspace but not the second, and $y$ is in the second subspace but not the first. Then I claim the $x+y$ can't be in either subspace, hence, can't be in their union; hence, the union is not closed under addition, so it's not a subspace.

So, let's prove the claim. If $x+y$ is in the first subspace, well, so is $x$, so $-x$ is also there, so $(x+y)+(-x)$ is there, but that's just $y$, which we know is not there. We've reached a contradiction on the assumption that $x+y$ was in the first subspace, so it can't be. Very similar reasoning shows it can't be in the second subspace, either, and we're done.

If $W_1$ and $W_2$ are subspaces then $W_1 \cup W_2$ is a subspace if and only if $W_1 \subset W_2$ or $W_2 \subset W_1$.

Proof:

($\Leftarrow$) This is the easy direction.

If $W_1 \subset W_2$ or $W_2 \subset W_1$ then we have $W_1 \cup W_2 = W_2$ or $W_1 \cup W_2 = W_1$, respectively. So $W_1 \cup W_2$ is a subspace as $W_1$ and $W_2$ are subspaces.

($\Rightarrow$) This is the harder direction and I give a direct proof.

Assuming $W_2 \not\subset W_1$, I'll show $W_1 \subset W_2$. Let $x \in W_1$ and $y \in W_2 - W_1$. So, by the definition of the union, we have $x \in W_1\cup W_2$ and $y \in W_1\cup W_2$. Therefore, as $W_1 \cup W_2$ is a subspace, $x + y \in W_1 \cup W_2$ which, again by the definition of the union, means that $x + y\in W_1$ or $x+y\in W_2$. If $x + y \in W_1$ then, as $W_1$ is a subspace, $y = (x + y) + (-x) \in W_1$ which is impossible as $y \in W_2-W_1$. So it must be that $x + y \in W_2$ in which case, as $W_2$ is a subspace, $x = (x + y) + (-y) \in W_2$. Therefore, as $x$ was arbitrary, $W_1 \subset W_2$ as desired. $_\Box$