Given a series of intermediate subspaces of $V$, prove that certain subspaces cannot exist iff $v_1, ..., v_n$ is a basis of $V$ Given a vector space $V$, we create a series of intermediate subspaces $V_1, ..., V_{n-1}$ such that $$ {0} = V_0 \subsetneq V_1 \subsetneq ... \subsetneq V_{n-1} \subsetneq V_n = V $$ We then create a list of vectors $v_1, ..., v_n \in V$ such that $v_i \in V_i$ while $v_i \notin V_{i-1}$. This problem asks to show that there does not exist a subspace $U \subseteq V$ such that $V_{i-1} \subsetneq U \subsetneq V_i$ for some $i$ if and only if the list $v_1, ..., v_n$ is a basis for the vector space $V$. 
My attempt at a solution for the 'if' part:
I first decided to prove that given $v_1, ..., v_n$ is a basis for $V$, then $U$ cannot exist. I first assumed that $U$ does exist. That implies a vector, say $v_u$, where $v_u \in U$ but $v_u \notin V_{i-1}$. We now have a list of vectors $v_1, ..., v_{i-1}, v_u, v_i, ..., v_n$ and I will choose to write them as a linear combination equal to $0$ in an attempt to show all vectors are linearly independent: $$ a_1v_1 + ... + a_{i-1}v_{i-1} + a_uv_u + a_iv_i + ... + a_nv_n = 0 $$ Now, I will assume that there is a vector that is in the span of other vectors, making the entire list of vectors dependent. If that is the case, I will look for the last nonzero coefficient $a_j$ in the linear combination above by examining $j$. 
Let's say that $j=u$. That is, $-a_uv_u = a_1v_1 + ... + a_{i-1}v_{i-1}$ and since $a_u$ is nonzero, $v_u \in span(v_1, ..., v_{i-1})$. This is a contradiction, as I stated that $v_u \notin V_{i-1}$ and the previous statement seems to suggest that $v_u \in V_{i-1}$.
So now I can say that the list from $v_1, ..., v_u$ must be linearly independent. Next, if $j > i$, then I can show that $v_j = \frac{1}{-a_j}(a_1v_1 + ... + a_uv_u + a_iv_i  + ... + a_{j-1}v_{j-1})$, which is also a contradiction since that would imply a vector $v_j \in V_{j-1}$. 
If $j < i$, we have a similar argument where $v_j = \frac{1}{-a_j}(a_1v_1 + ... a_{j-1}v_{j-1})$ implying $v_j \in V_{j-1}$ which is a contradiction with our starting premises.
However, I am stuck on the final case where $j = i$. If we tried to use the same argument, we get $v_i = \frac{1}{-a_j}(a_1v_1 + ... + a_uv_u)$ implying $v_i \in U$, but it might be the case that $v_i \in U$. I have been told that something should be modified in my choice of $v_u$, where $v_u \in U$ but $v_u \notin V_{i-1}$. My eventual goal is to show that the list $v_1, ..., v_u, ... v_n$ is linearly independent, which would mean that $v_u$ would have to be added to the initial basis $v_1, ..., v_n$ in order for it to still be a valid basis. However, the size of a basis of a vector space is fixed, so if $v_1, ..., v_n$ is already a basis for $V$, then $U$ cannot exist.
My attempt at a solution for the 'only if' part:
I will give a brief sketch for this one. I want to prove that if $U$ does not exist, then $v_1, ..., v_n$ must be a basis for $V$. First, if there is no $U$, then by the conditions of the problem only the vectors $v_1, ..., v_n$ exist such that $v_i \in V_i$ and $v_i \notin V_{i-1}$. Given this, I will prove (not listed here) that $v_1,..., v_n$ are linearly independent and we can always extend a linearly independent list of vectors to a basis. However, there are no more linearly independent vectors in $V$ due to $U$ not existing, and so $v_1, ..., v_n$ cannot be extended any more, which implies that that list is already a basis of $V$.
My questions:
1) How do I fix the $j=i$ case in the 'if' part of the proof?
2) Is my methodology for the 'only if' part sound? 
 A: The second part seems sound, given the sketch. I would need to see a little more detail before I was confident in the method, but nothing you said was wrong.
For the first part, I would go with what you're saying here:

My eventual goal is to show that the list $v_1,\ldots,v_u,\ldots,v_n$ is linearly independent

It's not hard to see that, regardless of whether $v_1, \ldots, v_n$ is a basis or not, if you make such selections, the resulting list is always linearly independent. This is because, for every $i$, you have that $v_1, \ldots, v_i \in V_i$, hence
$$\operatorname{span}(v_1, \ldots, v_i) \subseteq V_i,$$
and so if we take $v_{i+1} \notin V_i$, then $v_{i+1} \notin \operatorname{span}(v_1, \ldots, v_i)$, which is the condition required to extend a linearly independent list.
If you can insert a $U$ in there, the same argument shows $(v_1, \ldots, v_u, \ldots, v_n)$ is linearly independent, and therefore every basis for $V$ must consist of at least $n + 1$ vectors. This contradicts $v_1 \ldots, v_n$ being a basis. (This is not dissimilar to what David suggested.)
