Proving largest chain for subspaces of n-dimensional vector. In a course of mine, the following was given as an example of a chain (totally ordered set). 

Let $q \geq 1$ and suppose that $V$ is a $q$-dimensional vector space. Then let $A = \{W:W \text{ is a subspace of } V\}$, partially ordered by inclusion. 
Clearly $(A, \leq)$ is a poset, and if $B = \{v_{1},v_{2},\text{ ... }, v_{q}\}$ is a basis for $V$ and $W_{k} = \text{span}\{v_{1},v_{2},\text{ ... }, v_{k}\}$ where $k \in [1, q]$, then 
  $$
\mathcal{C} = \{\{0\}, W_{1}, \text{ ... }, W_{q}\} 
$$
  is a chain (or totally ordered set) in $A$. 

What I was wondering is if $\mathcal{C}$ is contained in any larger chain in $A$? 
My intuition makes me think this cannot be true but I'm having a hard time convincing myself of it formally. My thinking is that if we suppose there were some larger chain in which $\mathcal{C}$ was contained, it would have to contain a linearly dependent element with the same dimension as one of the basis vectors. Does it follow that these elements would not be orderable by inclusion? I don't really see how this would follow but it would complete the proof - I feel like I must be missing something. 
I haven't touched linear algebra in a few years so I'm a bit rusty in this department.
 A: There is no larger chain in $A$ that contains $\mathcal{C}$. This becomes clear by looking at the dimensions of the spaces in the chain; from the definition it follows that
$$\dim W_k=\dim\operatorname{span}(\{v_1,\ldots,v_k\})=k,$$
for $0\leq k\leq q$. Here I use that $W_0=\operatorname{span}(\varnothing)=\{0\}$.
It is clear that $W_0=\{0\}$ is minimal and $W_q=V$ is maximal with respect to inclusion. So if there is a chain in $A$ containing $\mathcal{C}$, then there exists a subspace $U\subset V$ such that $W_k\subsetneq U\subsetneq W_{k+1}$ for some $k\in\{0,\ldots,q\}$. It follows that
$$\dim U>\dim W_k=k\qquad\text{ and }\qquad \dim U<\dim W_{k+1}=k+1,$$
which is impossible.
A: Assume there was a larger chain. Then there'd be at least one additional subspace $U$ such that $W_k\subsetneq U\subsetneq W_{k+1}$ for some $k$ or $U\subsetneq \{0\}$ or $W_q\subsetneq U$.
The second is clearly impossible, and as $W_q=V$ so is the third case.
Let $u\in U\setminus W_k$ and write $u=a_1w_1+\ldots+a_kw_k+a_{k+1}w_{k+1}$. As $u\notin W_k$, we have $a_{k+1}\ne 0$. And as also $a_{k+1}w_{k+1}\in U$, we conclude $w_k\in U$, i.e., $W_{k+1}\subseteq U$.
