[proofplan]
We separate the tautological case $e \in A$ from the substantive case $e \notin A$. If $e$ lies in the closure of $A$ but not in $A$, then adjoining $e$ to a basis of $A$ creates dependence, and a minimal dependent subset of this dependent set gives the required circuit. Conversely, if such a circuit exists, then any basis of $A$ extending the independent set $C \setminus \{e\}$ becomes dependent after adjoining $e$, so $e$ cannot increase rank and therefore lies in the closure of $A$.
[/proofplan]
[step:Handle the elements already contained in $A$]
Let $M$ be a matroid on $E$, let $A \subset E$, and let $e \in E$. Let $r_M: \mathcal{P}(E) \to \mathbb{N} \cup \{0\}$ denote the rank function of $M$, so that $r_M(X)$ is the maximum cardinality of an independent subset of $X$ for each subset $X \subset E$. If $e \in A$, then $e \in \operatorname{cl}_M(A)$ by the extensivity property of closure, and the first alternative in the statement holds. Hence, for the rest of the proof, we may assume $e \notin A$.
[/step]
[step:Extract a circuit from a dependent extension of a basis of $A$]
Assume $e \notin A$ and $e \in \operatorname{cl}_M(A)$. Let $I \subset A$ be a basis of the restriction of $M$ to $A$; equivalently, $I$ is a maximal independent subset of $A$, and its cardinality is $r_M(A)$.
Since $e \in \operatorname{cl}_M(A)$ and $e \notin A$, adjoining $e$ does not increase rank:
\begin{align*}
r_M(A \cup \{e\}) = r_M(A).
\end{align*}
Because $I \subset A$, monotonicity gives $r_M(I \cup \{e\}) \le r_M(A \cup \{e\}) = r_M(A) = |I|$. Also $I \subset I \cup \{e\}$, so $r_M(I \cup \{e\}) \ge r_M(I) = |I|$. Therefore
\begin{align*}
r_M(I \cup \{e\}) = |I|.
\end{align*}
But $|I \cup \{e\}| = |I| + 1$ because $e \notin I$, so $I \cup \{e\}$ is dependent. By the circuit axiom for matroids, every dependent set contains a circuit; choose a circuit $C \subset I \cup \{e\}$.
The circuit $C$ must contain $e$. Indeed, if $e \notin C$, then $C \subset I$, contradicting the independence of $I$. Thus
\begin{align*}
e \in C \subset I \cup \{e\} \subset A \cup \{e\}.
\end{align*}
This proves the circuit alternative.
[guided]
Assume $e \notin A$ and $e \in \operatorname{cl}_M(A)$. The goal is to find a minimal dependent set whose only possible element outside $A$ is $e$. To do this, choose a basis $I \subset A$ of the restriction of $M$ to $A$. Thus $I$ is independent, $I \subset A$, and
\begin{align*}
|I| = r_M(A).
\end{align*}
Because $e$ is in the closure of $A$, adding $e$ to $A$ does not increase rank:
\begin{align*}
r_M(A \cup \{e\}) = r_M(A).
\end{align*}
Now compare the smaller set $I \cup \{e\}$ with $A \cup \{e\}$. Since $I \cup \{e\} \subset A \cup \{e\}$, rank monotonicity gives
\begin{align*}
r_M(I \cup \{e\}) \le r_M(A \cup \{e\}) = r_M(A) = |I|.
\end{align*}
On the other hand, $I \subset I \cup \{e\}$ and $I$ is independent, so
\begin{align*}
r_M(I \cup \{e\}) \ge r_M(I) = |I|.
\end{align*}
Combining the two inequalities yields
\begin{align*}
r_M(I \cup \{e\}) = |I|.
\end{align*}
Since $e \notin A$ and $I \subset A$, we have $e \notin I$, hence
\begin{align*}
|I \cup \{e\}| = |I| + 1.
\end{align*}
The rank of $I \cup \{e\}$ is therefore strictly smaller than its cardinality, so $I \cup \{e\}$ is dependent. A dependent set in a matroid contains a circuit, so choose a circuit $C \subset I \cup \{e\}$.
It remains to check that this circuit really uses the new element $e$. If $e \notin C$, then $C \subset I$. But $I$ is independent, and an independent set cannot contain a circuit. Therefore $e \in C$. Finally, because $I \subset A$, we have
\begin{align*}
e \in C \subset I \cup \{e\} \subset A \cup \{e\}.
\end{align*}
This is exactly the required circuit.
[/guided]
[/step]
[step:Use a circuit through $e$ to show that adjoining $e$ does not increase rank]
Conversely, assume that there exists a circuit $C$ of $M$ such that
\begin{align*}
e \in C \subset A \cup \{e\}.
\end{align*}
If $e \in A$, then $e \in \operatorname{cl}_M(A)$ by extensivity, so assume $e \notin A$. Then $C \setminus \{e\} \subset A$. Since $C$ is a circuit, $C \setminus \{e\}$ is independent.
The set $C \setminus \{e\}$ is independent in the restriction $M|A$, because $C \setminus \{e\} \subset A$ and independence is inherited by restrictions. By the extension axiom applied inside $M|A$, extend $C \setminus \{e\}$ to a basis $I \subset A$ of $M|A$. Then $I$ is independent, $C \setminus \{e\} \subset I$, and $|I| = r_M(A)$.
The set $I \cup \{e\}$ contains the circuit $C$, because $C \setminus \{e\} \subset I$ and $e \in I \cup \{e\}$. Hence $I \cup \{e\}$ is dependent. Since $I$ is independent and $e \notin I$, the set $I \cup \{e\}$ has cardinality $|I| + 1$ and contains an independent subset of cardinality $|I|$, namely $I$. Therefore
\begin{align*}
r_M(I \cup \{e\}) = |I|.
\end{align*}
We now use this particular dependent extension of a basis to control all of $A \cup \{e\}$. Since $I$ is a basis of $A$, every element of $A$ lies in $\operatorname{cl}_M(I)$; equivalently,
\begin{align*}
r_M(I \cup \{a\}) = r_M(I)
\end{align*}
for every $a \in A$. Also $e \in \operatorname{cl}_M(I)$, because
\begin{align*}
r_M(I \cup \{e\}) = r_M(I).
\end{align*}
By monotonicity of closure with respect to adjoining elements already in the closure, adding all elements of $A \cup \{e\}$ to $I$ does not increase rank. Hence
\begin{align*}
r_M(A \cup \{e\}) = r_M(I) = |I| = r_M(A).
\end{align*}
Therefore adjoining $e$ to $A$ does not increase rank, which is precisely $e \in \operatorname{cl}_M(A)$.
[guided]
Assume there is a circuit $C$ with
\begin{align*}
e \in C \subset A \cup \{e\}.
\end{align*}
If $e \in A$, then closure extensivity immediately gives $e \in \operatorname{cl}_M(A)$. Thus the only case needing proof is $e \notin A$.
Because $e \notin A$ and $C \subset A \cup \{e\}$, every element of $C$ except $e$ lies in $A$:
\begin{align*}
C \setminus \{e\} \subset A.
\end{align*}
Since $C$ is a circuit, it is minimally dependent. Therefore deleting any element from $C$ gives an independent set, so $C \setminus \{e\}$ is independent.
The set $C \setminus \{e\}$ is not only independent in $M$; it is independent in the restriction $M|A$, because it is contained in $A$ and restrictions preserve exactly the independent subsets lying in the restricted ground set. Therefore the extension axiom applies inside $M|A$: extend $C \setminus \{e\}$ to a basis $I \subset A$ of the restriction. This means that $I$ is independent, $C \setminus \{e\} \subset I$, and
\begin{align*}
|I| = r_M(A).
\end{align*}
Now look at $I \cup \{e\}$. Since $C \setminus \{e\} \subset I$ and $e \in I \cup \{e\}$, the whole circuit $C$ is contained in $I \cup \{e\}$. A set containing a dependent subset is dependent, so $I \cup \{e\}$ is dependent. Because $I$ itself is independent and $e \notin I$, the set $I \cup \{e\}$ has cardinality $|I| + 1$ and contains an independent subset of cardinality $|I|$. Thus the largest independent subsets of $I \cup \{e\}$ have size $|I|$, and
\begin{align*}
r_M(I \cup \{e\}) = |I| = r_M(I).
\end{align*}
This equality is the rank criterion saying that $e \in \operatorname{cl}_M(I)$.
Why does this imply the desired statement for $A$? Since $I$ is a basis of $A$, every element of $A$ is also in $\operatorname{cl}_M(I)$; in rank language, adding any $a \in A$ to $I$ does not increase the rank beyond $r_M(I)$. We have just proved the same for $e$. Hence every element of $A \cup \{e\}$ is already in the closure of $I$, so adjoining all of $A \cup \{e\}$ to $I$ does not increase rank:
\begin{align*}
r_M(A \cup \{e\}) = r_M(I).
\end{align*}
Since $I$ is a basis of $A$, $r_M(I) = |I| = r_M(A)$. Therefore
\begin{align*}
r_M(A \cup \{e\}) = r_M(A).
\end{align*}
By the rank characterization of matroid closure, this equality is exactly the assertion that $e \in \operatorname{cl}_M(A)$.
[/guided]
[/step]
[step:Combine the two implications]
The first step covers the case $e \in A$. The second step proves that, when $e \notin A$, membership $e \in \operatorname{cl}_M(A)$ produces a circuit $C$ with $e \in C \subset A \cup \{e\}$. The third step proves the converse implication from the existence of such a circuit to $e \in \operatorname{cl}_M(A)$. Hence, for every $A \subset E$ and every $e \in E$,
\begin{align*}
e \in \operatorname{cl}_M(A)
\end{align*}
if and only if either $e \in A$ or there exists a circuit $C$ of $M$ such that
\begin{align*}
e \in C \subset A \cup \{e\}.
\end{align*}
[/step]
