I'm not really sure what the proper procedure for this kind of question is. I was looking at this post which gives an (in my opinion) incomplete proof of the statement in the question title. I follow the entirety of the proof until the phrase
...the hereditary property+augmentation property of matroids also guarantees $(A\setminus\{a_i\})\cup \{\omega(a_i)\} \in \mathcal B$.
A year ago someone already asked for clarification regarding this very same line in the comments, but no one answered, so I'm not confident that my question will get answered if I just tack on another comment asking the exact same question. Thus why I'm creating a new post.
Everything makes sense to me up until this line. To my understanding, at this point in the proof we have shown that all of the following are bases: \begin{align} &\left\{ (A \cap B) \cup \{ \omega(a_l) , ..., \omega(a_1)\} \right\} &= C_{l+1}\\ &\left\{ (A \cap B) \cup \{ a_l\} \cup \{ \omega(a_{l - 1}) , ..., \omega(a_1)\} \right\} &= C_l\\ &\left\{ (A \cap B) \cup \{ a_l, a_{l-1}\} \cup \{ \omega(a_{l - 2}) , ..., \omega(a_1)\} \right\} &= C_{l-1} \\ &\vdots \\ &\left\{ (A \cap B) \cup \{ a_l, ..., a_2\} \cup \{ \omega(a_1)\} \right\} &= C_2 \\ &\left\{ (A \cap B) \cup \{ a_l, ..., a_2, a_1 \right\} &= A = C_1 \\ \end{align}
where $\{a_l, ... ,a_1\} = A \setminus (A \cap B)$, and the indexing is the same as used in the post I linked.
But now how do we show, for example, that $\{ (A \cap B)\cup \{a_l, ..., a_3, \omega(a_2), a_1 \} \}$ is a basis? All this post says is to "use the hereditary property + augmentation property." We know at this point that $\left\{ (A \cap B) \cup \{ a_l, ..., a_3\} \cup \{\omega(a_2), \omega(a_1)\} \right\} $ is a basis, and clearly we would love to be able to exchange $\omega(a_1)$ for $a_1$. But after playing around with it for a long time I see no way to force this exchange.
For example, we can't just add an element of $\left\{ (A \cap B) \cup \{ a_l, ..., a_2, a_1 \right\}$ to $\{ (A \cap B)\cup \{a_l, ..., a_3, \omega(a_2) \} \}$ because we might end up adding $a_2$ instead of $a_1$. Maybe I'm missing something very obvious, but I'm almost convinced a different method must be tried.
Thanks for the help!
I think the "proof" cited is wrong. Consider the graphic matroid over $[5]$ given by $\mathcal{C} = \{125,345,1234\}$ (two $C_3$'s -$125$ and $345$- with ${5}$ as a common edge). The bases of this matroid are
$$\mathcal{B} = \{123,124,134,135,145,234,235,245\}.$$
Let $A=C_1=134$, $B=235$, $a_1=4$, and $a_2=1$. Then, $C_1\backslash\{a_1\}=13$; to which I can add $2$ or $5$ from $B$. I pick $\omega(a_1)=2$. So, $C_2 = 123$. Next, $C_2\backslash\{a_2\}=23$; to which I can only add $5$ from $B$. So, $\omega(a_2)=5$ and $C_3=B$. But $A\backslash\{a_2\}\cup\{\omega(a_2)\}=345\not\in \mathcal{B}.$
The proof can be found in the paper "Comments on bases in dependence structures" by Richard A. Brualdi (1969). It is stated as Corollary 3.