Let $G$ be a graph and $A, B \subseteq V(G)$. Let $\mathfrak{F}$ be the set containing all $F \subseteq B$ that satisfy the condition: There are $|F|$ disjoint $a,b$-paths with $a\in A$ and $b\in F$ without inner vertices (vertices that aren't a or b) in $A\cup B$.
I want to show that $\mathfrak{F}$ is a matroid on $B$.
My attempt:
We can use induction on $|V(G)|$. The statement is trivial for $|V(G)| =0$ and $1$. Now assume the statement is true for graphs with less than $|V(G)|$ vertices and we want to show it for our G. It's not hard to see that the empty set is in $\mathfrak{F}$ and that subsets of sets in $\mathfrak{F}$ are in $\mathfrak{F}$ too. Now the hard part: We want to show that for $C, D \in \mathfrak{F}$ with $|C|<|D|$ there is a vertex $x\in D$ so that $ C\cup{} \{x\}\in \mathfrak{F}$. We can assume that there is at least one vertex in $C$ that is not in $D$ (otherwise $C$ would be a subset of $D$) and we can remove this vertex $v$ from G, then we can use our induction assumption on $C/\{v\}$ and $D$ in $G-v$. We add the vertex $x$ we get by doing this to $C$ and want to show that there are $|C|+1$ paths that satisfy the condition above.
There is a theorem that says: Let $A, B \subseteq V(G)$ and $S\subseteq V(G)$ be an $A,B$-seperator (that means every A,B-path contains an element from S), there are $k$ $S$-disjoint A,B-paths (that means that paths don't have common elements in S) iff there is no A,B-seperator $T\subseteq S$ with $|T|<k$. (this is a variation of Menger)
I want to apply this here on $A$, $C\cup\{x\}$ and G reduced on $|C|$ disjoint paths from C to A and $|D|$ disjoint paths from D to A. A seperator $T$ would have at least $|C|$ vertices but I don't know how to show that would also have $|C|+1$ vertices.
This type of matroid is known as the "$A\text{-}B$ gammoid over $G$" and it was proven 50 years ago that it is a contraction of a transversal matroid. For the special case of transversal matroids of any indexed family $\mathcal{F}=\{S_i\}_{i\in I}$ where no indice belongs to an indexed set we can take $V(G)=I\cup[\bigcup_{i\in I}S_i]$ and $E(G)=\bigcup_{i\in I}\{i\}\times S_i$ so that when $J$ is a basis of the transversal matroid $M(\mathcal{F})$ there is an injective map $f:J\to I$ thus $\mathcal{P}=\{(f(j),j):j\in J\}$ is a basis of a gammoid and vice versa all maximal matchings in $G$ recover the basis sets of $M(\mathcal{F})$ by collecting all vertices in each matching and then excluding those within $I$.
But in general for non-transversal gammoids look at a proof in the article:
https://www.sciencedirect.com/science/article/pii/0095895673900312