= Solution
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For the preliminary definitions, a block is $Re$ for a primitive central <idempotent> $e$, and a <module> belongs to it when $eM=M$. For a <semisimple algebra>, the <Artin–Wedderburn theorem> identifies its blocks with the factors $M_{n_i}(D_i)$ in its product decomposition. Each factor is a <block of a finite-dimensional algebra>.
We prove <Ext separation of finite-length modules>. The extension hypothesis is $\operatorname{Ext}^1(S,T)=\operatorname{Ext}^1(T,S)=0$ for simples in opposite sets. First, for a simple $S\in\mathcal C_1$ and a module $N$ of finite <composition length> with factors in $\mathcal C_2$, we have $\operatorname{Ext}^1(S,N)=0$. Induct on the length of $N$: for $0\to N'\to N\to T\to0$ with $T$ simple, the long exact sequence of the <Ext functor> contains
$$
\operatorname{Ext}^1(S,N')\longrightarrow\operatorname{Ext}^1(S,N)\longrightarrow\operatorname{Ext}^1(S,T),
$$
whose outer groups are zero. The same argument works with the two sets interchanged.
Now induct on the length of $M$, with $M=0$ immediate. Take a simple quotient in $0\to N\to M\to S\to0$. By induction, $N=N_1\oplus N_2$ with factors in the respective sets. Suppose $S\in\mathcal C_1$; the other case is symmetric. Quotienting by $N_1$ gives
$$
0\longrightarrow N_2\longrightarrow M/N_1\longrightarrow S\longrightarrow0.
$$
This splits by the preceding Ext vanishing. Let $U_1$ be the inverse image in $M$ of the chosen complementary copy of $S$. Then $U_1\cap N_2=0$, $U_1+N_2=M$, and $0\to N_1\to U_1\to S\to0$. Thus $U_1$ has only factors in $\mathcal C_1$, while $U_2=N_2$ has only factors in $\mathcal C_2$.
If two finite-length modules have factors in disjoint sets, any homomorphism between them is zero: a nonzero image would, by the <Jordan–Hölder theorem>, have a simple factor belonging to both sets. For any <submodule> $V$ of $M$ with factors in $\mathcal C_1$, projection onto $U_2$ is consequently zero, so $V\subset U_1$. The analogous argument applies to $U_2$. Hence
$$
\boxed{M=U_1\oplus U_2,\quad U_i\text{ is the unique largest submodule with factors in }\mathcal C_i}.
$$
In particular both summands are preserved by every endomorphism of $M$.
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