Q:

Composition Series; given a fin. dim. Module over an Algebra, any submodule is in a Composition Series

I have some problems with the following proposition and proof:
Proposition: Let $A$ be a $k$-algebra where $k$ is a field and let $V$ be a finite dimensional $A$-module [Note that this makes $V$ a vector space, defining for all $\lambda \in k, v \in V, \lambda v = (\lambda 1_{A})v]$
Suppose $L\subset V$ is a submodule of $V$. Then $V$ has a composition series which contains $L$.  
We later use this proposition in a proof of the Jordan-Holder theorem, so I would like to understand this result, but there is a statement which I can't get my head around in it's proof.  
Proof: We consider $V/L$ and $L$. If both are simple then $\{0\}\subset L \subset V$ is a composition series for $V$ containing $L$. If $L =\{0\}$ or $V$ then any composition series contains $L$. So we suppose not.
So one of $L, V/L$ isn't simple, say $V/L$ is not simple. So we can add in a submodule $L \subset W \subset V$ and consider the series $\{0\} \subset L \subset W \subset V$. If one of the composition factors isn't simple we add in a module in the suitable place...  
The proof then says that this process must end at a composition series for $V$ containing $L$ as $V$ is finite dimensional. But I am confused by this statement and it doesn't follow immediately to me.  
Some of my observations:
A submodule of a finite dimensional module doesn't have to be finite dimensional so we can't consider just the fact that basis elements must be added in at each stage surely?
Just because the module $V$ is finite dimensional over an algebra doesn't mean that $V$ is finite dimensional as a $k$-vector space:
For example set $k=\Bbb{Q}, A=\Bbb{R}, V=\Bbb{R}\;$ then clearly; $A=\Bbb{R}\,$ is a $k=\Bbb{Q}$-algebra.  $V$ is a finite dimensional $A$ module as $\{1\}$ is a basis for $\Bbb{R}$ when coefficients can be in $\Bbb{R}$, but $V=\Bbb{R}$ is not finite dimensional over $A=\Bbb{Q}$.
So why does the fact that $V$ is finite dimensional as a module mean that the process described above must end??  
Thanks in advance!!

A:

As you and Joppy observed in your comment above, the phrase let $V$ be a finite dimensional $A$-module means that $V$ is an $A$-module that is finite dimensional when considered as a $k$-vector space. In order to make clear how finite dimensionality is being used in the proof, I'd prefer to prove by induction on $n$ the statement 
Given an $A$-module $V$ of dimension $n$ over $k$, and a submodule $L$ of $V$, there exists a composition series for $V$ containing $L$.
Proof sketch: If $V$ is simple we're done so we may assume $V$ is not simple. If $L=V$ we choose a simple submodule $M$ of $V$ and apply induction to $V/M$. Otherwise $L<V$, so we choose a non-zero submodule $M<V$ containing $L$ (which we could take to be $L$ unless $L=0$) and apply induction to $M$ and $V/M$. 

