PROBLEM: Let $R$ be a ring with $1$ and $M$ be a unital $R$-module (i.e. $1x=x$). Let there for each submodule $M_1\neq M$ exist a submodule $M_2\neq M$, such that $M_1\cap M_2=\{0\}$. How can I prove, that $M$ is semisimple?
DEFINITIONS: A module $M$ is semisimple iff $\exists$ simple submodules $M_i\leq M$, such that $M=\bigoplus_{i\in I}M_i$. A module $M_i$ is simple iff it has no submodules (other than $\{0\}$ and $M$).
KNOWN FACTS: $M$ is semisimple $\Leftrightarrow$ $\exists$ simple submodules $M_i\leq M$, such that $M=\sum_{i\in I}M_i$ (the sum need not be direct) $\Leftrightarrow\forall\!M_1\!\leq\!M\;\exists M_2\!\leq\!M$ such that $M_1\oplus M_2=M$ (i.e. every submodule is a direct sumand).