Need help to prove the existence of set of linearly independent vectors. Let $S_{1}, S_{2}, \ldots S_{r}$ be r vector spaces such that 
$$
\dim\Bigl(\sum^{i_{l}}_{j=i_{1}}S_{i_{j}}\Bigr) \geq l
$$
for any $\{S_{i_{j}} , j = 1,2,\ldots l\} \subseteq \{S_{1},S_{2},\ldots, S_{r}\}$ for any $1 \leq l \leq r$ (i.e the sum of any l spaces among the collection has total dimension at least l, for each possible value of $l=1,2,\ldots r$. Thus each space is of dimension atleast 1, sum of any 2 has dimension atleast 2 and so on.)
Show that there exists at set of r linearly independent vectors, one from each $S_{i}$ i.e show that there exists a linearly independent set of vectors $\{v_{1},v_{2}, \ldots v_{r}\}$ such that $v_{i} \in S_{i}$.
My approach: I am using Proof by Contradiction. Here goes my attempt. Suppose there does not exist r linearly independent vectors $v_{i} \in S_{i}$, then any r vectors are linearly dependent, so we can express one of them as a linear combination of the others. 
$$
c_{1}v_{1} + c_{2}v_{2} + \ldots + c_{r}v_{r} = 0
$$
Let $v_{k}$ be the one to be expressed as linear combination of others, so we can write $v_{k}$ as
$v_{k} = c^{*}_{1}v_{1} + c^{*}_{2}v_{2} + \cdots + c^{*}_{k-1}v_{k-1} + c^{*}_{k+1}v_{k+1} + \cdots + c^{*}_{r}v_{r}$, where $c^{*}_{i} = \dfrac{-c_{i}}{c_{k}}$
Now consider sum of r vector spaces i.e $\sum^{r}_{j=1}S_{j}$. It can be written as,
$$
v_{1} + v_{2} + v_{3} + \cdots + v_{r} 
$$
such that $v_{i} \in S_{i}$. According to the condition given in the question, dimension of this should be atleast r. Substituting $v_{k}$ from above to get,
$$
v_{1} + v_{2} + \ldots + v_{k} + \cdots + v_{n}
$$
i.e
$$
(1+c^{*}_{1})v_{1} + (1+c^{*}_{2})v_{2} + \cdots + (1+c^{*}_{k-1})v_{k-1} + (1+c^{*}_{k+1})v_{k+1} + \cdots + (1+c^{*}_{n})v_{n}
$$
which is same as
$$
v_{1} + v_{2} +v_{3} + \cdots + v_{k-1} + v_{k+1} + \cdots + v_{r}
$$
Since $(1+c^{*}_{i})$ is the constant term and $(1+c^{*}_{i})$ when multiplied with $v_{i}$ can be written as $v_{i}$, because $v_{i}$ which in itself is a generalized term for vectors in $S_{i}$.
If we observe, this is the sum of r-1 vector spaces taken from the collection, i.e $v_{1},v_{2}, \cdots v_{k-1},v_{k+1},\cdots, v_{r}$.
So by condition given in the question the dimension of this is atleast r-1. So there rises the contradiction because according to the property given in the question sum of r vector spaces should be atleast r but we are getting atleast r-1. So a contradiction. Hence there exists a set of r linearly independent vectors, one from each $S_{i}$.
Is this approach correct ? If not, why and other approaches are also welcomed.
 A: Let $\underline{r} = \{1,\ldots,r\}$. For any non-empty subset $I \subseteq \underline{r}$, let $v_{Ijk} \in S_k$, for $j,k \in I$, such that
$$v_{Ij}:= \sum_{k\in I} v_{Ijk},$$ as $j$ runs in $I$, are linearly independent. This gives a collection of
$m = \sum_{k=1}^r k\binom{r}{k}$
vectors $v_{Ij}$. Indeed, there are $\binom{r}{k}$ subsets $I$ such that $|I| = k$, and for each such subset, there are $k$ vectors $v_{Ij}$, where $j \in I$.
Let $V = \sum_{i=1}^r S_i$. Define an $m\times r$ array of vectors in $V$ as follows. Index the set of rows by pairs $(I,j)$, where $I$ is a non-empty subset of $\underline{r}$ and $j \in I$. The row corresponding to $(I,j)$ should contain $v_{Ijk}$ in the $k$-th column, for $k \in I$, and contain the zero vectors at each column with index in $\underline{r}$ not belonging to $I$.
Let us denote this $m \times r$ array of vectors in $V$ by $A$. $A$ has the property that for any non-empty subset $I$ of $\underline{r}$, if we delete the columns of $A$ whose indices are in $\underline{r}$ but not in $I$, then the resulting submatrix $B$ has the property that one can choose from each row of $B$ a vector, such that the dimension of the span of the collection of chosen vectors is at least $|I|$.
Referring to A matroid problem inspired by a linear algebra problem, we have an $m \times r$ array $A$ of vectors in $V$ satisfying the hypotheses of the problem there. Then by the solution of that problem by A.B., we get that the column-rank of $A$ is $r$, which means that we can choose from each column $k$ of $A$ a vector, say $v_k$ (so that $v_k \in S_k$), in such a way that $v_1,\ldots,v_r$ are linearly independent. This solves positively the problem asked by the OP.
That was a non-trivial question! 
