Within my abstract algebra lecture we came across a theorem stating that for any ring $R$ and any $R$-module $M$ which is the sum of a family of simple submodules $(U_i)_{i\in I}$, i.e. $M=\sum_{i\in I}U_i$ we can show that for any simple submodule $N\le M$ we find a finite subset $J\subseteq I$ so that $N\cong U_j$ for every $j\in J$ (this is reasonably easy to prove) and $N\le\bigoplus_{j\in J}U_j$.
The proof goes as follows:
Seeing that $N$ is simple, it is given by $Ru$ for any $u\in N\setminus\{0\}$. Now since $u\in M$, there is a finite linear combination $u=\sum_{j\in J}u_j,\,u_j\in U_j\setminus\{0\},\, J$ finite. That implies the fact that $N=Ru\subseteq \sum_{j\in J}Ru_j=\sum_{j\in J}U_j$. Now we can look at the projection mapping $\pi_j\colon N\to U_j,\,ru\mapsto ru_j$ which is a non trivial homomorphism between simple modules, i.e. an isomorphism.
Thus far I can follow quite nicely. However, we do not address the problem of the above sum being direct. In general it is simply false to say that any sum of simple modules is direct already, so how come it is true in this example? Can anyone explain?
The answer to my question is reasonably straight forward if one actually uses the fact that the given $R$-module is semisimple: Because of that we can assume WLOG that the given sum is direct already, which makes the question trivial.