It's elementary if you know the basic properties: "$E(M)$ is a maximal essential extension of $M$", and "injective submodules are direct summands".
Suppose $E(M)$ is indecomposable. Let $A$ and $B$ are nonzero submodules of $M$ such that $A\cap B=0$. Then $E(A)$ is an injective submodule of $E(M)$, and hence it is a direct factor. The only possibilities are $E(M)$ and $0$. Since $A$ is nonzero, it is not the latter, so $E(A)=E(M)$. But this means that $A$ is essential in $E(M)$, and so $A\cap B\neq 0$, a contradiction.
Now suppose $0$ is irreducible. Let $C\oplus D=E(M)$ with $C\neq 0$. Now $(C\cap M)\cap (D\cap M)=0$, and since $M$ is essential in $E(M)$, $M\cap C\neq 0$. By irreducibility of $0$, $M\cap D=0$, but again because $M$ is essential, this amounts to $D=0$. Thus, $E(M)$ is indecomposable.