Let R be $K[x]$, the the ring of polynomials over K, and $A \in M_n(K)$. Then $K^2$ is a R-modul by using the Matrix A, with $p(x) \circ v=p(A)v$, where $v \in K^2$ and $p \in K[x]$.
Now I got show that $K^2$ for $A= \begin{pmatrix}0 & 1 \\ 0& 0\\ \end{pmatrix}$ is an indecomposable module. I'm not sure, but is it simply indecomposable because the eigenvalues are 0? Or have I to show more?
If you had instead chosen
$$ A= \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} $$
when defining $K^2$ as a $K[x]$-module, you would have obtained a decomposable one. So you need to say more than the fact that the above matrix has two eigenvalues $0$.
What is true is that a module gotten in this way will always decompose into eigenspaces, but sometimes it may decompose further (as in the case of the zero matrix), or it may not (as in the case of your matrix).
Here's a hint for your exercise: your module is actually cyclic, meaning there is a $v \in K^2$ such that any element $w \in K^2$ may be written as $p(x)v$ for some $p(x) \in K[x]$. Any cyclic module is irreducible.
