[proofplan]
We use the defining description of independent sets in the direct sum. Every independent subset of $A$ in $M$ splits uniquely into its part in $E_1$ and its part in $E_2$, and because the ground sets are disjoint, its cardinality is the sum of the two cardinalities. Maximizing over independent subsets of $A$ therefore separates into two independent maximization problems, one in $M_1$ and one in $M_2$.
[/proofplan]
[step:Split independent subsets of $A$ into their two direct-sum components]
Let $\mathcal{I}$ denote the independent-set family of $M$. For a subset $I \subset A$, the definition of the direct sum gives
\begin{align*}
I \in \mathcal{I} \iff I \cap E_1 \in \mathcal{I}_1 \text{ and } I \cap E_2 \in \mathcal{I}_2.
\end{align*}
Because $E_1 \cap E_2 = \varnothing$ and $E = E_1 \cup E_2$, every subset $I \subset E$ decomposes uniquely as
\begin{align*}
I = (I \cap E_1) \cup (I \cap E_2).
\end{align*}
Moreover, if $I \subset A$, then $I \cap E_1 \subset A \cap E_1$ and $I \cap E_2 \subset A \cap E_2$. Thus the independent subsets of $A$ in $M$ are precisely the sets of the form $I_1 \cup I_2$, where $I_1 \subset A \cap E_1$ is independent in $M_1$ and $I_2 \subset A \cap E_2$ is independent in $M_2$.
[/step]
[step:Separate the rank maximization into the two summands]
By definition of the rank function, $r_M(A)$ is the maximum cardinality of an independent subset of $A$ in $M$. Using the characterization from the previous step,
\begin{align*}
r_M(A)=\max\{|I_1 \cup I_2| : I_1 \subset A \cap E_1,\ I_1 \in \mathcal{I}_1,\ I_2 \subset A \cap E_2,\ I_2 \in \mathcal{I}_2\}.
\end{align*}
Since $E_1 \cap E_2 = \varnothing$, the sets $I_1$ and $I_2$ are disjoint, so
\begin{align*}
|I_1 \cup I_2|=|I_1|+|I_2|.
\end{align*}
Therefore
\begin{align*}
r_M(A)=\max\{|I_1|+|I_2| : I_1 \subset A \cap E_1,\ I_1 \in \mathcal{I}_1,\ I_2 \subset A \cap E_2,\ I_2 \in \mathcal{I}_2\}.
\end{align*}
The two choices are independent of one another: the condition on $I_1$ involves only $M_1$ and $A \cap E_1$, while the condition on $I_2$ involves only $M_2$ and $A \cap E_2$. Hence the maximum of the sum is the sum of the two maxima:
\begin{align*}
r_M(A)=\max\{|I_1| : I_1 \subset A \cap E_1,\ I_1 \in \mathcal{I}_1\}+\max\{|I_2| : I_2 \subset A \cap E_2,\ I_2 \in \mathcal{I}_2\}.
\end{align*}
By the definition of the rank functions $r_{M_1}$ and $r_{M_2}$, this is exactly
\begin{align*}
r_M(A)=r_{M_1}(A\cap E_1)+r_{M_2}(A\cap E_2).
\end{align*}
[guided]
We now turn the definition of the direct sum into a rank computation, starting from the decomposition itself. Let $\mathcal{I}$ denote the independent-set family of $M$, and let $\mathcal{I}_1$ and $\mathcal{I}_2$ denote the independent-set families of $M_1$ and $M_2$, respectively. For a subset $I \subset A$, the definition of the direct sum says
\begin{align*}
I \in \mathcal{I} \iff I \cap E_1 \in \mathcal{I}_1 \text{ and } I \cap E_2 \in \mathcal{I}_2.
\end{align*}
Because $E_1 \cap E_2 = \varnothing$ and the ground set of $M$ is $E_1 \cup E_2$, every subset $I \subset E_1 \cup E_2$ has the unique decomposition
\begin{align*}
I=(I \cap E_1)\cup(I \cap E_2).
\end{align*}
If $I \subset A$, then $I \cap E_1 \subset A \cap E_1$ and $I \cap E_2 \subset A \cap E_2$. Conversely, if $I_1 \subset A \cap E_1$ is independent in $M_1$ and $I_2 \subset A \cap E_2$ is independent in $M_2$, then $I_1 \cup I_2 \subset A$ and the direct-sum definition gives $I_1 \cup I_2 \in \mathcal{I}$, since $(I_1 \cup I_2) \cap E_1 = I_1$ and $(I_1 \cup I_2) \cap E_2 = I_2$. Thus the independent subsets of $A$ in $M$ are precisely the unions $I_1 \cup I_2$ with $I_1 \subset A \cap E_1$, $I_1 \in \mathcal{I}_1$, $I_2 \subset A \cap E_2$, and $I_2 \in \mathcal{I}_2$.
The rank $r_M(A)$ is defined by maximizing the size of an independent subset of $A$ in $M$. Using the correspondence just proved, we get
\begin{align*}
r_M(A)=\max\{|I_1 \cup I_2| : I_1 \subset A \cap E_1,\ I_1 \in \mathcal{I}_1,\ I_2 \subset A \cap E_2,\ I_2 \in \mathcal{I}_2\}.
\end{align*}
The disjointness hypothesis $E_1 \cap E_2 = \varnothing$ is used again in the cardinality computation. Since $I_1 \subset E_1$ and $I_2 \subset E_2$, the sets $I_1$ and $I_2$ are disjoint. Therefore the cardinality of their union is additive:
\begin{align*}
|I_1 \cup I_2|=|I_1|+|I_2|.
\end{align*}
Substituting this into the rank maximization gives
\begin{align*}
r_M(A)=\max\{|I_1|+|I_2| : I_1 \subset A \cap E_1,\ I_1 \in \mathcal{I}_1,\ I_2 \subset A \cap E_2,\ I_2 \in \mathcal{I}_2\}.
\end{align*}
The point of the direct sum is that there is no compatibility condition between $I_1$ and $I_2$. Once $I_1$ is independent in $M_1$ and $I_2$ is independent in $M_2$, their union is independent in $M$. Therefore maximizing $|I_1|+|I_2|$ separates into maximizing $|I_1|$ and maximizing $|I_2|$ independently:
\begin{align*}
r_M(A)=\max\{|I_1| : I_1 \subset A \cap E_1,\ I_1 \in \mathcal{I}_1\}+\max\{|I_2| : I_2 \subset A \cap E_2,\ I_2 \in \mathcal{I}_2\}.
\end{align*}
By the definition of rank in $M_1$, the first maximum is $r_{M_1}(A \cap E_1)$. By the definition of rank in $M_2$, the second maximum is $r_{M_2}(A \cap E_2)$. Hence
\begin{align*}
r_M(A)=r_{M_1}(A\cap E_1)+r_{M_2}(A\cap E_2).
\end{align*}
[/guided]
[/step]
