If union of n subspaces of V is a subspace of V, then one of the n subspaces must contain the other n-1 subspaces Help prove this? I can prove for n=2, but I'm stuck on proving it for general n. Thanks!
My proof for n=2
Forward direction: Consider A and B and A $\cup$ B is a subspace of V. We prove by contradiction by assuming $\exists$ x $\in$ A s.t x $\notin$ B and $\exists$ y $\notin$ s.t y $\in$ B. Consider x+y. Since x $\in$ A and y $\in$ B, if A$\cup$B is a subspace, we expect x+y $\in$ A$\cup$B. However, since y$\notin$A, then x+y $\notin$A because of closure under addition and multiplication of scalar. If, -x+x+y$\in$A then y$\in$A, but that is a contradiction. Thus x+y $\notin$ A and by similar logic, same for B. Since x+y $\notin$ A and x+y $\notin$ B, x+y $\notin$ A $\cup$ B which is a contradiction. Thus one of the subspaces must contain the other
Reverse direction: WLOG, A $\subset$ B so A $\cup$ B is equivalent to B. Since A and B are subspaces, they both have the 0 vector so A$\cup$B also does. For a1, a2 $\in$ A$\cup$B, a1, a2 $\in$ B, so a1+a2 $\in$A$\cup$B-->closed under addition. Consider b$\in$ A$\cup$B and scalar $\lambda$. b$\in$ B, and $\lambda$b $\in$ B so $\lambda$b $\in$ A$\cup$B.--> so closed under scalar multiplication. Thus A$\cup$B is a subspace.
Edit: Does it work that since I proved it for n=2, we can use induction to just look at V$\cup$W for dim(W)=1 as one of the comments pointed out? Doesn't that just reduce it to the 2 subspace case again? And since we know that the union is a subspace, is that too easy?

Let $U_1, U_2, \dots, U_n$ be subspaces of $V$ such that their union is also a subset of $V$.
We can suppose that $U_1 \nsubseteq U_2 \nsubseteq \dots \nsubseteq U_{n-1} \nsubseteq U_n \nsubseteq U_1$, because if $U_i \subseteq U_j$ for $i \ne j$ then we use induction on $n$ to conclude that there is $k \in \{1, \dots, n\} \setminus \{i\}$ such that $U_k$ contains all the other subsets.
Let $u_k \in U_k \setminus U_{k+1}$ for $k \in \{1, \dots, n\}$, where the indices of subspaces are modulo $n$. Suppose we have $N = n^2 - n + 1$ vectors in $\mathbb{F}^n$ such that any $n$ of then are linearly independent. Using them as coefficients, we construct the following
$$
x_1 = a_{11} u_1 + a_{12} u_2 + \dots + a_{1n} u_n\\
x_2 = a_{21} u_1 + a_{22} u_2 + \dots + a_{2n} u_n\\
\dots\\
x_N = a_{N1} u_1 + a_{N2} u_2 + \dots + a_{Nn} u_n\\
$$
As all $u_i$ are in $U_1 \cup U_2 \cup \dots \cup U_n$, so are all $x_i$. By the pigeonhole principle, there is $U_p$ that contains $n$ different $x_i$. WLOG, $x_1, \dots, x_n \in U_p$.
$$\begin{bmatrix}
a_{11} & a_{12} & \dots & a_{1n}\\
a_{21} & a_{22} & \dots & a_{2n}\\
 \dots &  \dots & \dots &  \dots\\
a_{n1} & a_{n2} & \dots & a_{nn}\\
\end{bmatrix}
\begin{bmatrix}u_1\\u_2\\\dots\\u_n\end{bmatrix}
= \begin{bmatrix}x_1\\x_2\\\dots\\x_n\end{bmatrix} 
$$
As we chose the vectors such that $x_1, \dots x_n$ are linearly independent, the matrix above is invertible. Therefore, each $u_i$ is written as a linear combination of the $x_i$, and so each $u_i \in U_p$. This is a contradiction, since $u_{p-1} \notin U_p$.
Therefore, we could not suppose that $U_1 \nsubseteq U_2 \nsubseteq \dots \nsubseteq U_{n-1} \nsubseteq U_n \nsubseteq U_1$, and so we conclude by induction that there is a subspace that contains all others.
