Question about direct sum of Noetherian modules is Noetherian Here is a corollary from Atiyah-Macdonald:

Question 1: The corollary states that finite direct sums of Noetherian modules are Noetherian. But they prove that countably infinite sums are Noetherian, right? (so they prove something stronger)
Question 2: I have come up with the following proof of the statement in the corollary, can you tell me if it's correct? Thank you:
Assume $M_i$ are Noetherian and let $(\bigoplus_i L_i)_n$ be an increasing sequence of submodules in $\bigoplus_i M_i$. Then in particular, $L_{in}$ is an increasing sequence in $M_i$ and hence stabilises, that is, for $n$ greater some $N_i$, $L_{in} = L_{in+1} = \dots $. Now set $N = \max_i N_i$. Then $(\bigoplus_i L_i)_n$ stabilises for $n> N$ and is equal to $\bigoplus_i L_i$, where $L_i = L_{iN_i}$.
This proves that finite direct sums of Noetherian modules are Noetherian so it's a bit weaker. But if it's correct it proves the corollary. 

As pointed out by others, the submodules of $M=M_1\oplus M_2$ are not necessarily direct sums of submodules of $M_1$ and $M_2$. Nevertheless, you always have the exact sequence 
$$0\rightarrow M_1\rightarrow M\rightarrow M_2\rightarrow 0$$
Then $M$ is Noetherian if (and only if) $M_1$ and $M_2$ are Noetherian. One direction is trivial (if M is Noetherian then $M_1$ and $M_2$ are Noetherian). I prove the other direction here:
Assume $M_1$ and $M_2$ are Noetherian. Given nested submodules $N_1\subset N_2\subset \cdots$ in $M$, we can see their images stabilize in $M_1$ and $M_2$. More precisely, the chain
$$(N_1\cap M_1)\subset (N_2\cap M_1)\subset\cdots$$ 
terminates in $M_1$, say at length $j_1$ and so does
$$(N_1+M_1)/M_1\subset (N_2+M_1)/M_1\subset\cdots$$ 
in $M_2$, say at length $j_2$. Set $j=\max\{j_1,j_2\}$ to get
$$(N_{j}\cap M_1)=(N_{j+1}\cap M_1)=\cdots$$
in $M_1$ and
$$(N_{j}+M_1)/M_1=(N_{j+1}+M_1)/M_1=\cdots$$
in $M_2$. But $N_j$'s are nested modules. Hence the above equalities can occur if and only if $N_j=N_{j+1}=\cdots$. To check this claim, pick $n\in N_{j+1}$. Then $m:=n'-n\in M_1$ for some $n'\in N_{j}$. But $m\in N_{j+1}$ as well. Hence $m\in N_{j+1}\cap M_1$, that is $m\in N_{j}\cap M_1$ by the first equality above. So $m\in N_j$, that is $n\in N_j$, giving us $N_{j+1}\subset N_j$.
So the chain $N_1\subset N_2\subset ...$ terminates in $M$.
