[proofplan]
We prove the contrapositive of the desired containment statement by assuming that the set obtained from $C_1 \cup C_2$ after deleting $e$ is independent. The rank of this set then forces the rank of $C_1 \cup C_2$ to have exactly one possible value. [Submodularity of the matroid rank function](/theorems/5819) then implies that $C_1 \cap C_2$ is dependent, contradicting the fact that it is a proper subset of a circuit. Therefore the deleted union is dependent, and a minimal dependent subset of it is the required circuit.
[/proofplan]
[step:Assume the deleted union is independent and compute its rank]
Define the set $X \subset E$ by
\begin{align*}
X := (C_1 \cup C_2) \setminus \{e\}.
\end{align*}
Suppose, for contradiction, that $X$ is independent. Since $M$ is finite and $X$ is independent, the definition of the rank function gives
\begin{align*}
r(X) = |X| = |C_1 \cup C_2| - 1.
\end{align*}
Define $Y \subset E$ by
\begin{align*}
Y := C_1 \cup C_2.
\end{align*}
Then $Y = X \cup \{e\}$. We use the following one-element rank bound, which follows directly from the definition of rank: every independent subset of $Y$ either lies in $X$ or becomes an independent subset of $X$ after deleting $e$. Therefore
\begin{align*}
r(X) \le r(Y) \le r(X) + 1.
\end{align*}
The alternative $r(Y) = r(X) + 1 = |Y|$ is impossible, because $r(Y) = |Y|$ would mean that $Y$ is independent, and then its subset $C_1$ would be independent, contradicting that $C_1$ is a circuit. Hence
\begin{align*}
r(Y) = r(X) = |C_1 \cup C_2| - 1.
\end{align*}
[guided]
Define
\begin{align*}
X := (C_1 \cup C_2) \setminus \{e\}
\end{align*}
and assume, toward a contradiction, that $X$ is independent. The point of this assumption is that it turns the rank of $X$ into an exact cardinality: by the definition of rank in a finite matroid,
\begin{align*}
r(X) = |X|.
\end{align*}
Since $e \in C_1 \cap C_2$, deleting $e$ from $C_1 \cup C_2$ removes exactly one element, so
\begin{align*}
r(X) = |X| = |C_1 \cup C_2| - 1.
\end{align*}
Now define
\begin{align*}
Y := C_1 \cup C_2.
\end{align*}
Then $Y = X \cup \{e\}$. Adding one element to a set can increase matroid rank by at most one: an independent subset of $Y$ either avoids $e$, in which case it lies in $X$, or contains $e$, in which case deleting $e$ gives an independent subset of $X$ with one fewer element. Therefore
\begin{align*}
r(X) \le r(Y) \le r(X) + 1.
\end{align*}
There are only two possible values for $r(Y)$. We rule out the larger one. If
\begin{align*}
r(Y) = r(X) + 1,
\end{align*}
then using $r(X) = |Y| - 1$ gives
\begin{align*}
r(Y) = |Y|.
\end{align*}
For a finite matroid, $r(Y) = |Y|$ means that $Y$ itself is independent. But independence is hereditary, so every subset of $Y$ would also be independent. Since $C_1 \subset Y$, this would make $C_1$ independent, contradicting that $C_1$ is a circuit, hence dependent. Thus the larger value is impossible, and we must have
\begin{align*}
r(Y) = r(X) = |C_1 \cup C_2| - 1.
\end{align*}
[/guided]
[/step]
[step:Use rank submodularity to force the intersection to be dependent]
Since $C_1$ and $C_2$ are circuits, each is minimally dependent. Thus every proper subset of $C_i$ is independent, and in particular
\begin{align*}
r(C_i) = |C_i| - 1
\end{align*}
for $i \in \{1,2\}$. We use the submodularity property of the matroid rank function: for any subsets $A, B \subset E$, one has $r(A \cup B) + r(A \cap B) \le r(A) + r(B)$. Applying this property with $A := C_1$ and $B := C_2$ gives
\begin{align*}
r(C_1 \cup C_2) + r(C_1 \cap C_2) \le r(C_1) + r(C_2).
\end{align*}
Substituting the rank values gives
\begin{align*}
|C_1 \cup C_2| - 1 + r(C_1 \cap C_2) \le |C_1| + |C_2| - 2.
\end{align*}
Using the finite-set identity
\begin{align*}
|C_1| + |C_2| = |C_1 \cup C_2| + |C_1 \cap C_2|
\end{align*}
we obtain
\begin{align*}
r(C_1 \cap C_2) \le |C_1 \cap C_2| - 1.
\end{align*}
Therefore $C_1 \cap C_2$ is dependent.
[/step]
[step:Contradict minimal dependence of the two circuits]
Because $C_1$ and $C_2$ are distinct circuits, neither can properly contain the other: if, for instance, $C_1 \subsetneq C_2$, then $C_1$ would be a dependent proper subset of the circuit $C_2$, contradicting minimal dependence of $C_2$. Hence $C_1 \cap C_2$ is a proper subset of both $C_1$ and $C_2$. Since every proper subset of a circuit is independent, $C_1 \cap C_2$ is independent. This contradicts the conclusion of the previous step that $C_1 \cap C_2$ is dependent.
[/step]
[step:Extract a circuit contained in the deleted union]
The contradiction shows that $X = (C_1 \cup C_2) \setminus \{e\}$ is dependent. Since $M$ is finite, among all dependent subsets of $X$ there exists one of minimal cardinality; denote such a subset by $C_3 \subset X$. By minimality, $C_3$ is dependent and every proper subset of $C_3$ is independent, so $C_3$ is a circuit of $M$. Since $C_3 \subset X$, we have
\begin{align*}
C_3 \subset (C_1 \cup C_2) \setminus \{e\}.
\end{align*}
This is the required circuit.
[/step]
