[proofplan]
We verify the independent-set axioms for a matroid directly. The empty set satisfies every block quota, and any subset of a quota-satisfying set still satisfies the same quotas. For augmentation, if two independent sets $I,J \in \mathcal{I}$ satisfy $|I|<|J|$, then some block $E_j$ contains strictly fewer elements of $I$ than of $J$; choosing an element of $J$ from that block but not in $I$ preserves all quotas when added to $I$.
[/proofplan]
[step:Verify that the empty set satisfies every partition quota]
Since $\varnothing \cap E_j = \varnothing$ for every $j \in \{1,\dots,m\}$, we have
\begin{align*}
|\varnothing \cap E_j| = 0 \leq b_j.
\end{align*}
Thus $\varnothing \in \mathcal{I}$.
[/step]
[step:Show that subsets of independent sets remain independent]
Let $I \in \mathcal{I}$, and let $A \subset I$. For every $j \in \{1,\dots,m\}$, the inclusion $A \subset I$ gives
\begin{align*}
A \cap E_j \subset I \cap E_j.
\end{align*}
Taking cardinalities of finite sets gives
\begin{align*}
|A \cap E_j| \leq |I \cap E_j| \leq b_j,
\end{align*}
where the last inequality holds because $I \in \mathcal{I}$. Hence $A \in \mathcal{I}$.
[/step]
[step:Find a block where the larger independent set has more elements]
Let $I,J \in \mathcal{I}$ satisfy $|I|<|J|$. Suppose, for contradiction, that
\begin{align*}
|I \cap E_j| \geq |J \cap E_j|
\end{align*}
for every $j \in \{1,\dots,m\}$. Since $E_1,\dots,E_m$ are pairwise disjoint and have union $E$, every subset $A \subset E$ decomposes as the disjoint union
\begin{align*}
A = \bigcup_{j=1}^m (A \cap E_j).
\end{align*}
Therefore
\begin{align*}
|I| = \sum_{j=1}^m |I \cap E_j|
\geq \sum_{j=1}^m |J \cap E_j|
= |J|,
\end{align*}
contradicting $|I|<|J|$. Hence there exists $j_0 \in \{1,\dots,m\}$ such that
\begin{align*}
|I \cap E_{j_0}| < |J \cap E_{j_0}|.
\end{align*}
[guided]
Let $I,J \in \mathcal{I}$ satisfy $|I|<|J|$. We need to locate a part of the partition where $J$ has more elements than $I$; that is the only kind of part from which we can take a new element of $J$ and add it to $I$.
Assume instead that no such part exists. This means that for every $j \in \{1,\dots,m\}$,
\begin{align*}
|I \cap E_j| \geq |J \cap E_j|.
\end{align*}
Because $E_1,\dots,E_m$ are pairwise disjoint and their union is $E$, each subset $A \subset E$ is partitioned by its intersections with the blocks:
\begin{align*}
A = \bigcup_{j=1}^m (A \cap E_j),
\end{align*}
and this union is disjoint. Therefore the cardinality of $A$ is the sum of its blockwise cardinalities:
\begin{align*}
|A| = \sum_{j=1}^m |A \cap E_j|.
\end{align*}
Applying this first to $A=I$ and then to $A=J$, and using the assumed blockwise inequalities, gives
\begin{align*}
|I| = \sum_{j=1}^m |I \cap E_j|
\geq \sum_{j=1}^m |J \cap E_j|
= |J|.
\end{align*}
This contradicts the hypothesis $|I|<|J|$. Therefore the assumption was false, so there exists $j_0 \in \{1,\dots,m\}$ such that
\begin{align*}
|I \cap E_{j_0}| < |J \cap E_{j_0}|.
\end{align*}
[/guided]
[/step]
[step:Add an element from that block without violating any quota]
Choose $j_0 \in \{1,\dots,m\}$ such that
\begin{align*}
|I \cap E_{j_0}| < |J \cap E_{j_0}|.
\end{align*}
Then $(J \cap E_{j_0}) \setminus I$ is nonempty; choose
\begin{align*}
e \in (J \cap E_{j_0}) \setminus I.
\end{align*}
Since $J \in \mathcal{I}$, we have
\begin{align*}
|J \cap E_{j_0}| \leq b_{j_0}.
\end{align*}
The strict inequality between finite cardinalities gives
\begin{align*}
|I \cap E_{j_0}| + 1 \leq |J \cap E_{j_0}| \leq b_{j_0}.
\end{align*}
Now verify every quota for $I \cup \{e\}$. For the block $E_{j_0}$,
\begin{align*}
|(I \cup \{e\}) \cap E_{j_0}|
= |I \cap E_{j_0}| + 1
\leq b_{j_0},
\end{align*}
because $e \in E_{j_0}$ and $e \notin I$. If $j \neq j_0$, then $e \notin E_j$ because the sets $E_1,\dots,E_m$ are pairwise disjoint, so
\begin{align*}
|(I \cup \{e\}) \cap E_j|
= |I \cap E_j|
\leq b_j.
\end{align*}
Thus $I \cup \{e\} \in \mathcal{I}$.
[/step]
[step:Conclude the matroid axioms]
The family $\mathcal{I}$ contains $\varnothing$, is closed under passing to subsets, and satisfies the augmentation axiom: whenever $I,J \in \mathcal{I}$ with $|I|<|J|$, there exists $e \in J \setminus I$ such that $I \cup \{e\} \in \mathcal{I}$. Therefore $(E,\mathcal{I})$ is a matroid.
[/step]
