Looking for a good counterargument against vector space decomposition. How do I see that I cannot write $\mathbb{R}^n = \bigcup_{\text{all possible }M} \operatorname{span}(M)$, where $M$ runs over the subsets with $n-1$ elements in it of the set of vectors $N=\{a_1,\ldots,a_n,\ldots,a_m\} \in \mathbb{R}^n$, where the total dimension of the span of all of them is $n$.

Let $V$ be a vector space over an infinite field $F$, and $V_1, \ldots, V_n$ proper subspaces.  Then I claim $\bigcup _j V_j$ is not a vector space.
For each $k$ let $u_k$ be a vector not in $V_k$. We then inductively find
vectors $w_k$ not in $\bigcup_{j \le k} V_j$.  Namely, if $w_k \notin \bigcup_{j \le k} V_j$, and $u_{k+1} \notin V_{k+1}$, consider $f(t) = t u_{k+1} + (1-t) w_k$ for scalars $t$.  If this was in $V_j$ for two different values of $t$, say $t_1 \ne t_2$, then it would be in $V_j$ for all $t$, because
$$f(t) = \dfrac{t - t_1}{t_2 - t_1} f(t_2) + \dfrac{t - t_2}{t_1 - t_2} f(t_1)$$
This is not the case for any $j \in \{1,\ldots,k+1\}$ because $f(0) \notin V_{k+1}$ and $f(1) \notin V_j$ for $j \le k$.  So there are at most $k+1$
values of the scalar $t$ for which $f(t) \in \bigcup_{j \le k+1} V_j$, and
infinitely many for which it is not.  
