Using matrix theory to solve this problem I'm sorry that I couldn't find a better title for this. I was wondering if my solution is valid for the following problem, or if I've made some mistake.
Problem: Let $N=\{a_1, \dots, a_n\}$ be a finite set with $n$ elements. Let $M_1, \dots, M_{n+1}$ be $n+1$ non-empty subsets of $N$. Show that one can always choose two non-empty index sets $I$ and $J$ in $\{1,\dots,n+1\}$ such that $I \cap J=\emptyset$ and 
$$\cup_{i\in I} M_i = \cup_{j \in J} M_j.$$
What I've done so far: Consider, for every subset $M_l$, a vector $X_l$ such that $x_k=1$ if $a_k \in M$ and $0$ otherwise. All $X_l \not= 0$.
The problem statement is equivalent to saying that one can always choose two non-empty index sets $I$ and $J$ in $\{1,\dots,n+1\}$ such that $I \cap J=\emptyset$ and $\{X_i : i\in I\}$ and $\{X_j : j\in J\}$ span the same space. 
Assume the opposite is true, i.e. for every non-empty $I$ and $J$ the following holds: $\{X_i : i\in I\}$ and $\{X_j : j\in J\}$ span different spaces (or $I \cup J=\emptyset$, which will never happen since $I$ and $J$ are non-empty). This thus means that all $X_k$, $k=1,\dots,n+1$ are linearly independent, which leads to a contradiction since $n+1$ vectors of length $n$ cannot be linearly independent. 
 A: No, your proof does not work. That $\{X_i: i \in I\}$ and $\{X_j: j \in J\}$ span different spaces does not imply that $\{X_i: i \in I \} \cup \{X_j: j \in J\}$ is a linearly independent set of vectors. 
For example: $\left\{\begin{pmatrix} 1 \\ 0\end{pmatrix}, \begin{pmatrix} 0 \\ 1 \end{pmatrix}\right\}$ and $\left\{\begin{pmatrix} 1 \\ 1 \end{pmatrix}\right\}$ span different spaces, but if you take all three vectors together, they are not linearly independent.
Take your vectors $X_k$. Since you have $n+1$ vectors in an $n$-dimensional space, there must exist scalars $\alpha_k$, with at least two of them non-zero, such that:
$$\sum_{k=1} \alpha_k X_k = 0.$$
Now, if $\alpha_k$ is negative, set $\beta_k = |\alpha_k|$ and put $k$ in the set $J$. If $\alpha_k$ is positive, put $k$ in the set $I$. If $\alpha_k$ is zero, do not put $k$ in any set. You can now write:
$$\sum_{i \in I} \alpha_i X_i = \sum_{j \in J} \beta_j X_j$$
and here the set of non-zero coordinates must be the same on both sides, implying the result:
$$\bigcup_{i \in I} M_i = \bigcup_{j \in J} M_j.$$
A stronger result
An interesting strengthening of this result is that if you have $n+2$ distinct $M_i$-sets, you can find disjoint non-empty $I,J$ such that 
$$\bigcap_{i \in I} M_i = \bigcap_{j \in J} M_j$$
also holds.
See Lindström, Another theorem on families of sets, Ars Combinatoria 35, 123-124 or Jukna, Extremal Combinatorics, Springer Verlag, section 14.2.
