Unions of subspaces So I need to prove that for the union of $n$ subspaces to be a subspace,  each subspace must be a subset of another one of the subspaces.  My thought process so far is that I need to prove that it is possible for a union of $n$ subspaces to be a subspace space (which I proved). and that I need to prove that if an element is in some subspace of $w_j$ and not in the union, then the union is not a subspace.  (I can't seem to figure out how to prove this part).  Any hints or tips, I have been working on it for a while now, I feel like i am missing an obvious solution.
 A: Let $\mathcal F$ be a finite family of subspaces of $V$, a vector field over an infinite field $F$.
We want to prove $\bigcup\limits_{W\in \mathcal F}W\neq \bigoplus\limits_{W\in \mathcal F}W$.
We shall prove if $\mathcal J$ is a finite family of proper subspaces of $\bigoplus\limits_{W\in F}W$ then $\bigcup\limits_{W\in \mathcal J}W\neq \bigoplus\limits_{W\in \mathcal F}W$. By induction over $|\mathcal J|$, base case is trivial. Note we can take $\mathcal J$ so no subspace contains any other.
Inductive step:
Take $W\in \mathcal J$, and take $w\in W$ so that it is not in any of the other subspaces (possible by inductive step). Take a nonzero vector $v\not\in W$, then the set $A=\{fw+v|f\in F\}$ is infinite since $F$ is infinite.
Moreover any $U\in \mathcal J$ contains at most one element of $A$. Clearly $W$ contains none of them. If another subspace contained $fw+v$ and $gw+v$ it would contain $(f-g)w$ and so it would contain $w$, a contradiction.
Hence every subspace contains at most one element of $A$, since $A$ is infinite then this means some elements of $A$ are not in $\bigcup\limits_{W\in \mathcal J}W$. Therefore $\bigcup\limits_{W\in \mathcal J}W\neq \bigoplus\limits_{W\in \mathcal F}W$ as desired.
A: Let's assume the negative of the statement, each subspace must be a subset of another subspace (except for the "biggest" one). 
Then there exists a subspace $W_i$, where $W_i \not \subset W_j$ for all $j \neq i$ and $W_j \not \subset W_i$ for at least one j, denoted by $W_k$. If this last condition didn't hold then $W_i$ would just be the "biggest" set as noted in your comment. 
Now the first condition says that for each pair $W_i, W_j$, there is an element, call it $w_{ij}$ such that $w_{ij} \in W_i$ and $w_{ij} \not \in W_j$. The second says that there is a subspace $W_k$ which has an element, call it $w_k$, which is not in $W_i$.
Now note that $\sum_{j \neq i} w_{ij} \in W_i$ since $W_i$ is a subspace. However, $w_{ij} \not \in W_j$ for all $j \neq i$, and so $\sum_{j \neq i} w_{ij} \not \in W_j$ for all $j \neq i$. (If the sum was in $w_j$, then $w_{ij}$ would be in $W_j$ because $W_j$ is a subspace, which is a contradiction of our definition of $w_{ij}$). Therefore $\sum_{j \neq i} w_{ij} \not \in \cup \,W_j$. 
Now take $v = w_k + \sum_{j \neq i} w_{ij}$. This is not in $W_i$, because this would imply that $w_k$ is in $W_i$, which contradicts our earlier assumption. It is also not in $\cup \, W_j$, because this would imply $\sum_{j \neq i} w_{ij}$ would be in $\cup \, W_j$ (this assumes that $\cup \, W_j$ is a subspace, which must be true otherwise we are done anyway). So we have shown that v is not in $\cup \, W_j$ for all j, which means that the union of the subspaces is not closed under addition, which means it is not a subspace.  
