Let $G = (V,E)$ be an undirected graph. Set $M_k(G) = (E,S)$ where $$S = \{F ∪M | F ⊆ E,(V,F) \;\text{acyclic},M ⊆ E,|M| ≤ k\}.$$
Show that $M_k(G)$ is the set of independent sets of a matroid! In particular, the set of spanning forests of an undirected graph $G = (V,E)$ is the set of independent sets of a matroid.
I really need help with this one, I have no idea where to start.
I think the OP means to say that $M_k(G)=(E,S)$ is a matroid on the ground set $E$ with $S$ being the family of independent sets. I assume here that $G$ is a finite graph (so $V$ and $E$ are finite sets). I don't know how to deal with infinite matroids. The second part of the task is simply showing that $M_0(G)$ is a matroid, which is implied by the first part.
From here, there are three checkpoints. For the first one, it is easy to see that $\emptyset \in S$, since $F=\emptyset$ and $M=\emptyset$ gives $\emptyset=F\cup M\in S$.
For the hereditary property, let $F\cup M\in S$ and $T\subseteq F\cup M$. Write $T_F=T\cap F$ and $T_M=T\cap M$. Then, the edge induced subgraph $G[T_F]$ of $G$ is a forest, and $|T_M|\leq |M|\leq k$. So, $T=T_F\cup T_M$ is in $S$.
For the augmentation property, let $F_1\cup M_1,F_2\cup M_2\in S$ be such that $|F_1\cup M_1|<|F_2\cup M_2|$. WLOG, assume that $M_1\cap F_1=\emptyset$ and $M_2\cap F_2=\emptyset$. In the case $|M_2|>|M_1|$, things are pretty easy. Pick an element $m\in (F_2\cup M_2)\setminus (F_1\cup M_1)$. We have $$\big|M_1\cup\{m\}\big| =|M_1|+1\leq |M_2|\leq k.$$ So, $(F_1\cup M_1)\cup\{m\}=F_1\cup \big(M_1\cup\{m\}\big)$ is in $S$. If $|M_2|\leq |M_1|$, then we must have $$|F_2|-|F_1|\geq \big(|F_2|-|F_1|\big)+\big(|M_2|-|M_1|\big)=\big(|F_2|+|M_2|\big)-\big(|F_1|+|M_1|\big).$$ Because $F_1\cap M_1=F_2\cap M_2=\emptyset$, $$|F_2|-|F_1|\geq |F_2\cup M_2|-|F_1\cup M_1|>0.$$ Therefore, the forest $F_1$ has more connected components than $F_2$ (recalling that the number of connected components in a forest on $n$ vertices and $m$ edges is $n-m$). Therefore, there exists a connected component $C$ of $F_2$ with vertices from at least two connected components of $F_1$. Therefore, there is an edge $e$ of $C$ that connects two distinct connected components of $F_1$. Note that $F_1\cup\{e\}$ is still a forest. So, $(F_1\cup M_1)\cup \{e\}=(F_1\cup\{e\})\cup M_1\in S$.