[proofplan] We define a closure operator by sending each subset $A \subseteq E$ to the smallest member of $\mathcal F$ containing $A$. The intersection axiom gives extensivity, monotonicity, and idempotence. The partition axiom for covers of a flat gives the exchange axiom for this closure operator. We then construct independent sets from this closure operator, verify the matroid independence axioms, and finally check that the closed sets of the resulting matroid are precisely the members of $\mathcal F$. [/proofplan]

[step:Define the closure operator from the flat family] For each subset $A \subseteq E$, define $\operatorname{cl}: \mathcal P(E) \to \mathcal P(E)$ by \begin{align*} \operatorname{cl}(A) := \bigcap \{F \in \mathcal F : A \subseteq F\}. \end{align*} The indexing family is nonempty because $E \in \mathcal F$ and $A \subseteq E$. Since $E$ is finite, $\mathcal F$ is finite, so this is an intersection of members of $\mathcal F$; by the intersection axiom, $\operatorname{cl}(A) \in \mathcal F$. The map $\operatorname{cl}$ is extensive because $A$ is contained in every member of $\mathcal F$ containing $A$, hence $A \subseteq \operatorname{cl}(A)$. It is monotone because if $A \subseteq B$, then every member of $\mathcal F$ containing $B$ also contains $A$, so the intersection defining $\operatorname{cl}(A)$ is taken over a larger family and therefore \begin{align*} \operatorname{cl}(A) \subseteq \operatorname{cl}(B). \end{align*} It is idempotent because $\operatorname{cl}(A) \in \mathcal F$ and contains $A$, so it is the smallest member of $\mathcal F$ containing $A$; therefore \begin{align*} \operatorname{cl}(\operatorname{cl}(A)) = \operatorname{cl}(A). \end{align*} [/step]

[step:Use the cover partition axiom to prove exchange]Let $A \subseteq E$, let $e,f \in E$, and suppose \begin{align*} f \in \operatorname{cl}(A \cup \{e\}) \setminus \operatorname{cl}(A). \end{align*} Set $F := \operatorname{cl}(A)$. Since $f \notin F$, the hypothesis also implies $e \notin F$; otherwise monotonicity and idempotence would give \begin{align*} \operatorname{cl}(A \cup \{e\}) \subseteq \operatorname{cl}(F) = F, \end{align*} contradicting $f \notin F$. By the partition axiom applied to $F$, there is a unique member $G_e \in \mathcal C(F)$ such that $e \in G_e \setminus F$. Since $G_e \in \mathcal F$ contains $F \cup \{e\}$, the minimality of $\operatorname{cl}(F \cup \{e\})$ gives \begin{align*} \operatorname{cl}(F \cup \{e\}) \subseteq G_e. \end{align*} The set $\operatorname{cl}(F \cup \{e\})$ is a member of $\mathcal F$, properly contains $F$, and is contained in the cover $G_e$. By minimality of $G_e$ over $F$, we have \begin{align*} \operatorname{cl}(F \cup \{e\}) = G_e. \end{align*} Since $A \subseteq F$ and $F = \operatorname{cl}(A)$, idempotence gives \begin{align*} \operatorname{cl}(A \cup \{e\}) = \operatorname{cl}(F \cup \{e\}) = G_e. \end{align*} Thus $f \in G_e \setminus F$. By uniqueness in the partition, the unique cover $G_f \in \mathcal C(F)$ with $f \in G_f \setminus F$ is $G_f = G_e$. Repeating the preceding argument with $f$ in place of $e$ yields \begin{align*} \operatorname{cl}(F \cup \{f\}) = G_f = G_e. \end{align*} Therefore $e \in G_e = \operatorname{cl}(F \cup \{f\})$. Since $F = \operatorname{cl}(A)$, this gives \begin{align*} e \in \operatorname{cl}(A \cup \{f\}). \end{align*} Hence $\operatorname{cl}$ satisfies the exchange axiom.[/step]

[guided]We prove the exchange axiom directly from the partition axiom. Fix $A \subseteq E$ and elements $e,f \in E$ such that \begin{align*} f \in \operatorname{cl}(A \cup \{e\}) \setminus \operatorname{cl}(A). \end{align*} Define the flat \begin{align*} F := \operatorname{cl}(A). \end{align*} The point is to understand what happens immediately above the flat $F$. Since $f \notin F$, we must also have $e \notin F$. Indeed, if $e \in F$, then $A \cup \{e\} \subseteq F$, so monotonicity gives \begin{align*} \operatorname{cl}(A \cup \{e\}) \subseteq \operatorname{cl}(F). \end{align*} Because $F$ is closed under our operator, $\operatorname{cl}(F)=F$, which would force $f \in F$, contradicting the assumption. Now apply the cover partition axiom to the flat $F$. The family of sets $G \setminus F$, where $G$ ranges over the minimal members of $\mathcal F$ properly containing $F$, partitions $E \setminus F$. Since $e \in E \setminus F$, there is a unique cover $G_e \in \mathcal C(F)$ with \begin{align*} e \in G_e \setminus F. \end{align*} Because $G_e$ is a member of $\mathcal F$ containing $F \cup \{e\}$, the definition of closure as the smallest flat containing a set gives \begin{align*} \operatorname{cl}(F \cup \{e\}) \subseteq G_e. \end{align*} The set $\operatorname{cl}(F \cup \{e\})$ belongs to $\mathcal F$, contains $F$, and contains $e$, so it properly contains $F$. Since $G_e$ is minimal among members of $\mathcal F$ properly containing $F$, the containment above must be equality: \begin{align*} \operatorname{cl}(F \cup \{e\}) = G_e. \end{align*} Because $F = \operatorname{cl}(A)$, adjoining $e$ after closing $A$ does not change the generated flat: \begin{align*} \operatorname{cl}(A \cup \{e\}) = \operatorname{cl}(F \cup \{e\}) = G_e. \end{align*} The assumed membership of $f$ therefore says $f \in G_e$. Since $f \notin F$, this is really $f \in G_e \setminus F$. The partition axiom gives uniqueness of the cover whose difference from $F$ contains $f$, so the cover $G_f$ associated to $f$ must equal $G_e$. Applying the same closure-minimality argument to $f$ gives \begin{align*} \operatorname{cl}(F \cup \{f\}) = G_f = G_e. \end{align*} Since $e \in G_e$, we conclude \begin{align*} e \in \operatorname{cl}(F \cup \{f\}) = \operatorname{cl}(A \cup \{f\}). \end{align*} This is exactly the exchange axiom.[/guided]

[step:Define independent sets from the closure operator] Define \begin{align*} \mathcal I := \{I \subseteq E : x \notin \operatorname{cl}(I \setminus \{x\}) \text{ for every } x \in I\}. \end{align*} The empty set belongs to $\mathcal I$ because the defining condition is vacuous. If $I \in \mathcal I$ and $J \subseteq I$, then $J \in \mathcal I$. Indeed, for $x \in J$, we have $J \setminus \{x\} \subseteq I \setminus \{x\}$, so monotonicity gives \begin{align*} \operatorname{cl}(J \setminus \{x\}) \subseteq \operatorname{cl}(I \setminus \{x\}). \end{align*} Since $x \notin \operatorname{cl}(I \setminus \{x\})$, it follows that $x \notin \operatorname{cl}(J \setminus \{x\})$. Thus $J \in \mathcal I$. [/step]

[step:Prove the exchange property for independent sets] We first record the replacement principle forced by closure exchange. Let $Z,Y \subseteq E$ and let $x \in E$ satisfy \begin{align*} x \in \operatorname{cl}(Z \cup Y) \setminus \operatorname{cl}(Z). \end{align*} Choose a subset $Y_0 \subseteq Y$ minimal by inclusion such that \begin{align*} x \in \operatorname{cl}(Z \cup Y_0). \end{align*} Then $Y_0$ is nonempty. For any $y \in Y_0$, minimality gives \begin{align*} x \notin \operatorname{cl}(Z \cup (Y_0 \setminus \{y\})). \end{align*} Applying closure exchange with base $Z \cup (Y_0 \setminus \{y\})$, adjoining element $y$, and exchanged element $x$, we obtain \begin{align*} y \in \operatorname{cl}(Z \cup (Y_0 \setminus \{y\}) \cup \{x\}). \end{align*} Consequently, replacing $y$ by $x$ does not change the closure of $Z \cup Y$: \begin{align*} \operatorname{cl}(Z \cup Y) = \operatorname{cl}(Z \cup (Y \setminus \{y\}) \cup \{x\}). \end{align*} Now let $X,Y \subseteq E$ with $X \in \mathcal I$ and $X \subseteq \operatorname{cl}(Y)$. We prove $|X| \le |Y|$. List $X$ as $X=\{x_1,\dots,x_m\}$. Starting from $Y$, the replacement principle successively replaces one element outside $\{x_1,\dots,x_{k-1}\}$ by $x_k$, because independence of $X$ gives \begin{align*} x_k \notin \operatorname{cl}(\{x_1,\dots,x_{k-1}\}). \end{align*} After $k$ replacements, the resulting set has the same cardinality as $Y$, has the same closure as $Y$, and contains $\{x_1,\dots,x_k\}$. This process can continue through $k=m$ only if $m \le |Y|$. Hence $|X| \le |Y|$. Now take $I,J \in \mathcal I$ with $|I| < |J|$. If every $y \in J \setminus I$ satisfied $y \in \operatorname{cl}(I)$, then $J \subseteq \operatorname{cl}(I)$. Applying the preceding paragraph with $X=J$ and $Y=I$ would give $|J| \le |I|$, contradicting $|I|<|J|$. Therefore there exists $y \in J \setminus I$ such that \begin{align*} y \notin \operatorname{cl}(I). \end{align*} Then $I \cup \{y\} \in \mathcal I$. For $y$ this is the displayed condition. For $x \in I$, if \begin{align*} x \in \operatorname{cl}((I \setminus \{x\}) \cup \{y\}), \end{align*} then exchange with base $I \setminus \{x\}$ would imply \begin{align*} y \in \operatorname{cl}(I), \end{align*} contradicting the choice of $y$. Thus $I \cup \{y\}$ is independent. This proves the independent-set augmentation axiom. [/step]

[step:Construct the matroid and identify its flats] The preceding steps show that $\mathcal I$ satisfies the three independence axioms: $\varnothing \in \mathcal I$, heredity holds, and augmentation holds. Therefore $\mathcal I$ is the family of independent sets of a matroid $M$ on ground set $E$. It remains to identify the flats of $M$. Let $\operatorname{cl}_M$ denote the matroid closure operator determined by $\mathcal I$. For any $A \subseteq E$, choose a maximal independent subset $B \subseteq A$. Such a subset exists because $E$ is finite. Maximality of $B$ in $A$ implies $A \subseteq \operatorname{cl}(B)$: if $a \in A \setminus \operatorname{cl}(B)$, then the exchange argument used above shows $B \cup \{a\}$ is independent, contradicting maximality. Hence \begin{align*} \operatorname{cl}(A) = \operatorname{cl}(B). \end{align*} For $x \in E$, the set $B \cup \{x\}$ is independent exactly when $x \notin \operatorname{cl}(B)$. Therefore adjoining $x$ to $A$ increases the maximum size of an independent subset exactly when $x \notin \operatorname{cl}(A)$. By the definition of matroid closure, this proves \begin{align*} \operatorname{cl}_M(A) = \operatorname{cl}(A) \end{align*} for every $A \subseteq E$. Thus the flats of $M$ are exactly the fixed points of $\operatorname{cl}$. Since every member of $\mathcal F$ is fixed by construction, and every fixed point of $\operatorname{cl}$ belongs to $\mathcal F$, the collection of flats of $M$ is precisely $\mathcal F$. This completes the proof. [/step]