[proofplan]
We verify the independent-set axioms for $\mathcal I$, viewed as the candidate independent sets of a matroid. The empty-set and hereditary axioms follow directly from the definition, so the only substantive point is exchange. For exchange, we prove the standard local circuit-exchange lemma, then use it to show that an independent set of smaller size can be enlarged by an element from the larger independent set. Finally, the circuits of the resulting matroid are exactly the prescribed members of $\mathcal C$ because dependence means containing one of them and the members of $\mathcal C$ are inclusion-minimal among themselves.
[/proofplan]
[step:Verify the empty-set and hereditary independent-set axioms]
Since every member of $\mathcal C$ is nonempty, the empty set contains no member of $\mathcal C$. Hence $\varnothing \in \mathcal I$.
Let $I \in \mathcal I$, and let $I_0 \subset I$. If there were a circuit $C \in \mathcal C$ with $C \subset I_0$, then $C \subset I$, contradicting $I \in \mathcal I$. Therefore $I_0 \in \mathcal I$. Thus $\mathcal I$ is nonempty and hereditary.
[/step]
[step:Prove that exchanging one element from a fundamental circuit preserves independence]
We first prove the local exchange fact that replaces the invalid Hall-neighbourhood argument.
[claim:Single-circuit exchange]
Let $I\in\mathcal I$, let $e\in E\setminus I$, and suppose $C\in\mathcal C$ satisfies
\begin{align*}
e\in C\subset I\cup\{e\}.
\end{align*}
If $x\in C\setminus\{e\}$, then
\begin{align*}
(I\setminus\{x\})\cup\{e\}\in\mathcal I.
\end{align*}
[/claim]
[proof]
Define the set
\begin{align*}
I_x:=(I\setminus\{x\})\cup\{e\}.
\end{align*}
Assume, toward a contradiction, that $I_x\notin\mathcal I$. By the definition of $\mathcal I$, there exists a circuit $D\in\mathcal C$ with
\begin{align*}
D\subset I_x.
\end{align*}
Since $I\setminus\{x\}\subset I$ and $I\in\mathcal I$, the circuit $D$ cannot be contained in $I\setminus\{x\}$. Hence $e\in D$.
The circuits $C$ and $D$ are distinct, because $x\in C$ while $x\notin I_x$ and therefore $x\notin D$. Applying the stated circuit elimination axiom to the distinct circuits $C$ and $D$ at the common element $e$, we obtain a circuit $C'\in\mathcal C$ such that
\begin{align*}
C'\subset (C\cup D)\setminus\{e\}.
\end{align*}
Now $C\setminus\{e\}\subset I$ and $D\setminus\{e\}\subset I\setminus\{x\}\subset I$, so
\begin{align*}
(C\cup D)\setminus\{e\}\subset I.
\end{align*}
Thus $C'\subset I$, contradicting $I\in\mathcal I$. Therefore $I_x\in\mathcal I$.
[/proof]
[guided]
The useful local fact is this: if adding $e$ to an independent set $I$ creates a circuit $C$, then any old element $x$ of that circuit can be removed in exchange for $e$. We prove this directly from the stated circuit elimination axiom.
Define
\begin{align*}
I_x:=(I\setminus\{x\})\cup\{e\}.
\end{align*}
Suppose $I_x$ were dependent. Since dependence here means containing a member of $\mathcal C$, there is a circuit $D\in\mathcal C$ with
\begin{align*}
D\subset I_x.
\end{align*}
The set $I\setminus\{x\}$ is a subset of the independent set $I$, so it contains no circuit. Therefore $D$ cannot lie entirely in $I\setminus\{x\}$, and the only remaining element of $I_x$ is $e$. Hence $e\in D$.
Now compare the two circuits $C$ and $D$. They both contain $e$. They are distinct because $x\in C$ by choice, while $x\notin I_x$ and therefore $x\notin D$. The circuit elimination axiom applies to $C$ and $D$ at the element $e$, giving a circuit $C'\in\mathcal C$ such that
\begin{align*}
C'\subset (C\cup D)\setminus\{e\}.
\end{align*}
But after removing $e$, every element of $C$ lies in $I$, because $C\subset I\cup\{e\}$, and every element of $D$ lies in $I\setminus\{x\}\subset I$, because $D\subset I_x$ and $e$ has been removed. Therefore
\begin{align*}
(C\cup D)\setminus\{e\}\subset I.
\end{align*}
Thus $C'\subset I$, which contradicts the assumption $I\in\mathcal I$. This contradiction proves
\begin{align*}
(I\setminus\{x\})\cup\{e\}\in\mathcal I.
\end{align*}
[/guided]
[/step]
[step:Prove the exchange axiom by replacing elements until an element of $J\setminus I$ can be added]
Let $I,J\in\mathcal I$ satisfy $|I|<|J|$. We prove that there exists $e\in J\setminus I$ such that $I\cup\{e\}\in\mathcal I$.
Assume, toward a contradiction, that
\begin{align*}
I\cup\{e\}\notin\mathcal I
\end{align*}
for every $e\in J\setminus I$. Define a finite sequence of independent sets as follows. Let
\begin{align*}
S_0:=I.
\end{align*}
As long as there exists $e\in J\setminus S_k$ with $S_k\cup\{e\}\notin\mathcal I$, choose such an element $e_k$, choose a circuit $C_k\in\mathcal C$ with
\begin{align*}
C_k\subset S_k\cup\{e_k\},
\end{align*}
and choose an element
\begin{align*}
x_k\in (C_k\setminus\{e_k\})\setminus J.
\end{align*}
The element $x_k$ exists because $J\in\mathcal I$, $e_k\in J$, and otherwise $C_k\subset J$. Set
\begin{align*}
S_{k+1}:=(S_k\setminus\{x_k\})\cup\{e_k\}.
\end{align*}
By the single-circuit exchange claim, $S_{k+1}\in\mathcal I$.
At each step, $|S_{k+1}|=|S_k|=|I|$, and
\begin{align*}
|S_{k+1}\cap J|=|S_k\cap J|+1,
\end{align*}
because $x_k\notin J$ and $e_k\in J\setminus S_k$. Since $S_k$ is finite, this process can occur at most $|I\setminus J|$ times. Hence it terminates at an independent set $S_m\subset I\cup J$ with $|S_m|=|I|$ and with the property that every element $e\in J\setminus S_m$ satisfies
\begin{align*}
S_m\cup\{e\}\in\mathcal I.
\end{align*}
Because $|S_m|=|I|<|J|$, the set $J\setminus S_m$ is nonempty. Choose $e\in J\setminus S_m$. Then $S_m\cup\{e\}\in\mathcal I$. Since $S_m$ is obtained from $I$ by replacing elements of $I\setminus J$ with elements of $J\setminus I$, every element of $I\setminus S_m$ lies in $I\setminus J$, and every element of $S_m\setminus I$ lies in $J\setminus I$. Therefore
\begin{align*}
I\cup\{e\}\subset S_m\cup\{e\}\cup (I\setminus S_m).
\end{align*}
This containment alone does not give independence, so we instead reverse the replacements one at a time. Let $S_m,S_{m-1},\dots,S_0=I$ be the reverse sequence. If $S_r\cup\{e\}\in\mathcal I$ for some $1\le r\le m$, then
\begin{align*}
S_{r-1}\cup\{e\}=(S_r\setminus\{e_{r-1}\})\cup\{x_{r-1},e\}.
\end{align*}
If $S_{r-1}\cup\{e\}$ were dependent, it would contain a circuit $D\in\mathcal C$. This circuit must contain $x_{r-1}$, since $S_r\cup\{e\}$ is independent and $S_{r-1}\cup\{e\}$ differs from it only by replacing $e_{r-1}$ with $x_{r-1}$. Applying circuit elimination to $D$ and $C_{r-1}$ at $x_{r-1}$ gives a circuit contained in $S_r\cup\{e\}$, contradicting $S_r\cup\{e\}\in\mathcal I$. Hence $S_{r-1}\cup\{e\}\in\mathcal I$.
Descending induction gives $S_0\cup\{e\}=I\cup\{e\}\in\mathcal I$, contradicting the assumption that no element of $J\setminus I$ can be added to $I$. Therefore there exists $e\in J\setminus I$ with
\begin{align*}
I\cup\{e\}\in\mathcal I.
\end{align*}
This proves the exchange axiom.
[/step]
[step:Identify the circuits of the resulting matroid with the prescribed family]
The preceding steps show that $\mathcal I$ satisfies the empty-set, hereditary, and exchange axioms. Hence $\mathcal I$ is the independent-set system of a matroid on $E$.
It remains to identify its circuits. By definition, a dependent set for this matroid is a subset $A\subset E$ with $A\notin\mathcal I$, equivalently a subset containing at least one member of $\mathcal C$. Thus every $C\in\mathcal C$ is dependent. Since no member of $\mathcal C$ properly contains another member of $\mathcal C$, no proper subset of $C$ contains a member of $\mathcal C$. Hence every proper subset of $C$ is independent, so $C$ is a circuit of the matroid.
Conversely, let $D\subset E$ be a circuit of the matroid. Then $D\notin\mathcal I$, so there exists $C\in\mathcal C$ with $C\subset D$. Since $D$ is minimally dependent, no proper subset of $D$ is dependent. Therefore $C=D$. Thus every matroid circuit is a member of $\mathcal C$, and the circuits are exactly the prescribed sets.
[/step]
