Let $R$ be a right noetherian ring. Then every injective envelope is a direct sum of indecomposable modules. I got stuck trying to understand the proof of this lemma. 
All the modules we're considering are right modules over $R$, where $R$ is right noetherian. By indecomposable module, I mean a module that can't be expressed as inner direct sum of two non-trivial modules.
1
The claim is true for the injective envelope of a cyclic module $mR$:
$mR \simeq \frac{R}{ker \varphi}$ for $\varphi: r \mapsto mr$, and factors of $R_R$ are noetherian, so $mR$ is noetherian.
If there was no way to describe $E(mR)$ (the injective envelope of $mR$) as a direct sum of indecomposable modules, we could define a module that is an infinite direct product of nontrivial modules $F_i$, such that this inner direct sum is also a submodule of $E(mR)$. Using the fact that $mR$ is essential in $E(mR)$, we get that the interesection $mR \cap F_i$ is nontrivial, and so we get a contradiction with $mR$ being noetherian.
Therefore $E(mR)$ is a direct sum of indecomposable modules.
2
Notice that for any submodule $N$ of $E(M)$, the corresponding injection induces an injection from $E(N)$ to $E(M)$. $E(N)$ injective, and so it's a direct summand of $E(M)$.
3
Here is where I got stuck - 
Define:
$I = \{ F \subseteq E(M) |  F$ is a direct summand of $E(M)$, $F$ is a direct sum of indecomposable modules$\}$
$I$ contains the empty set module, so it's not empty. 
We will use Zorn's lemma to show that this set has a maximal element. If we take a non-decreasing chain $F_\alpha$, the union can be described as a direct sum of injective modules - this is because we know that $F_2 = F_1 \oplus A_1$ using injectivity (a direct summand of $E(M)$ is injective) - therefore we get that $F_n = F_1 \oplus \bigoplus_{i<n} A_i $.
How do we show, that the union is also a direct sum of indecomposable modules?
How would the proof continue after that?
I think I'm missing something. Maybe I could try and show that if I have $A \oplus B = C$ for $C$ a module that can be expressed as a direct sum of indecomposable modules, then $A$ and $B$ can also be expressed as a direct sum of indecomposable modules. 
I'm also not sure, whether it's maybe possible for there to exist a way of expressing a module as a direct sum of indecomposable modules, and yet there existing an expression of it by modules that aren't indecomposable, and that can 'always' be further decomposed?
Needless to say, I'm a confused about this topic overall too, and advice on what excercises would be useful to do, would be appreciated.
 A: Here is a way I thought of proving 3:
Instead of simply looking at the set $I$ as the set defined in the question (a subset of $\mathscr P(E(M))$, we look at $J = \{ A \subseteq \mathscr P(E(M))$ | $\sum_{a \in A} a$ is a direct summand of $E(M)$, all the elements of $A$ are indecomposable modules, and  $ a \in A$ are linearly independent $ \}$, $J$ being a non-empty subset of $\mathscr P (\mathscr P(E(M)))$ (in contains the empty set).
Because $a \in A$ are linearly independent (in the sense that a sum of elements from $\sum_{a \in A} a$ is $0$, iff all the elements are $0$), it makes sense to consider $\sum_{a \in A} a$ to be a inner direct product of the modules from $A$ and denote it as $\oplus_{a \in A} a$ instead.
We use Zorn's lemma on $J$, with the partial ordering on $J$ being the one induced by inclusion. 
Consider $A = \cup A_i$, $A_i \in J$ being a nondecreasing system - we will try to show that $A \in J$. Any $a \in A$ is in some $A_i$, and thus it is an indecomposable module.  We would like to show that $\sum_{a \in A} a$ is a direct sum - take a finite amount of elements $v_1,..,v_n$ from $\sum_{a \in A} a$, such that $v_1 + .. +v_n=0$ and they are all in distinct $a \in A$ modules. They are necessarily contained in some $\sum_{a \in A_i} a$, with $a \in A_i$ being linearly independent modules, thus each $v_i$ is $0$, and so $a \in A$ are also linearly independent modules. 
Any $a \in A$ is contained some $A_i$, and so it is a direct summand of $\sum_{a \in A_i} a$, which is a direct summand of $E(M)$, which is injective - thus, any $a \in A$ is a direct summand of $E(M)$, and so it is also injective. Because of this, $\sum_{a \in A} a$ is a direct sum of injective modules, which is injective, because they are all modules over a noetherian ring $R$.
Therefore conditions of Zorn's lemma are satisfied, and there exists a maximal element in $J$, let us call it $A$, and let us call $B=\sum_{a \in A} a = \oplus_{a \in A} a$.
$B$ is a direct summand of $E(M)$, so $E(M)=B\oplus G$ for some $G$. Let $m \in G\cap M$, $G$ is injective as well, so we can show that $E(mR) \subseteq G$, and by 1 we know that $E(mR)$ can be written as a direct sum of indecomposable modules. But by maximality, $B=B \oplus E(mR)$, thus $E(mR)=0$, so $m=0$ and $G \cap M = 0$. Because $M$ is essential in $E(M)$, it follows that $G =0$, and so $B=E(M)$, as we wanted to show.
Also, an answer to the little question I wrote at the end - yes, there is a module that can be expressed as a direct sum of indecomposable modules, but such that I can keep decomposing it (non-trivially) infinitely - take for example $\mathbb{R}^{(\omega)}$ as a vector space over $\mathbb{R}$ - this is a direct sum of $\mathbb{R}$ modules - this expression is a direct sum of indecomposable modules, but I can also express it as $\mathbb{R}^{(\omega)}= \mathbb{R} \oplus  M_1 = \mathbb{R} \oplus \mathbb{R}  \oplus M_2 $ etc.  - which gives me an "infinite decomposition$.
