Let $M:=\begin{pmatrix} \Bbb C \\ \Bbb C \end{pmatrix}$ be a left $R:=\begin{pmatrix} \Bbb C & 0 \\ \Bbb C & \Bbb C \end{pmatrix}$-module.
We would like to $M$ is not completely reducible as an $R$-module.
But the only thing that I can think to use is the definition: An $R$-module $M$ is called completely reducible, if $\forall N \leq M$, $\exists L \leq M$ such that $M=N\oplus L$.
Could you please help?
Thanks in advance.
Well, the answer nearly forces itself upon you: $M$ is $2$ dimensional over $\mathbb C$, so any proper submodule would have to be $1$ dimensional, and couldn't be a summand.
With that in mind you quickly come to the realization that $N=\begin{bmatrix}0\\ \mathbb C\end{bmatrix}$ is a submodule of $M$. In fact, if $\begin{bmatrix}a\\ b\end{bmatrix}$ is any element in $M$, then $\begin{bmatrix}0 & 0\\ 1 & 0\end{bmatrix}\begin{bmatrix}a\\ b\end{bmatrix}=\begin{bmatrix}0\\ a\end{bmatrix}$ is in the same submodule as $\begin{bmatrix}a\\ b\end{bmatrix}$.
So suppose $N'+N=M$. $N'$ has to have an element which is nonzero in the top entry (why?) But then it must have nontrivial intersection with $N'$. Therefore $N$ never has a complement.