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Source. Lemma 3, printed p. 151 (published PDF).

Statement. For an independent set $I$ and any element $e$, $I\cup\{e\}$ contains at most one circuit.

The printed lemma is this uniqueness assertion alone. The proof of Theorem 1c on p. 151 also uses, citing Lemma 3, that when $I\cup\{e\}$ is dependent, removing any element of its unique circuit makes it independent; that consequence is proved in the last paragraph below.

Proof. Every circuit in $I\cup\{e\}$ contains $e$, since $I$ is independent. Suppose two distinct circuits $C_1,C_2$ exist, and choose a counterexample with $|I|$ as small as possible. Neither circuit contains the other, so choose $a\in C_1\setminus C_2$ and $b\in C_2\setminus C_1$. Both belong to $I$ and are distinct.

The set

$$D=(I\cup\{e\})\setminus\{a,b\}$$

is independent. Otherwise it contains a circuit $C_3$, necessarily containing $e$. Then $(I\setminus\{a\})\cup\{e\}$ contains both $C_2$ and $C_3$. They are distinct because $b\in C_2$ but $b\notin C_3$. This contradicts minimality of $|I|$.

Both $I$ and $D$ are maximal independent in $I\cup\{e\}$. The former cannot take $e$; the latter cannot take $a$ because $D\cup\{a\}$ contains $C_1$, and cannot take $b$ because $D\cup\{b\}$ contains $C_2$. Their sizes, $|I|$ and $|I|-1$, are different, contradicting the matroid axiom. This proves uniqueness.

If $I\cup\{e\}$ is dependent, finiteness supplies a circuit $C$. After deleting $x\in C$, any remaining dependence would contain another circuit of $I\cup\{e\}$, different from $C$ because it omits $x$. Uniqueness excludes this. $\square$