Given is an undirected connected graph $ G=(V, E) $
(a) Justify: there is a spanning tree $ T $ for $ G, $ i. e. a subgraph $ T=\left(V, E^{\prime}\right) $ so that $ E^{\prime} \subseteq E $ and $ T $ is a tree.
(b) Let $ T_{1}=\left(V, E_{1}\right), T_{2}=\left(V, E_{2}\right) $ be spanning trees of $ G $.
Show: $ \forall e_{1} \in E_{1} \exists e_{2} \in E_{2}:\left(V,\left(E_{1} \backslash\left\{e_{1}\right\}\right) \cup\left\{e_{2}\right\}\right) $ and $ \left(V,\left(E_{2} \backslash\left\{e_{2}\right\}\right) \cup\left\{e_{1}\right\}\right) $ are spanning trees of $ G $.
How do I show this?
Not sure Why you kept the title 'Minimum Spanning Tree Proof, but I will try to answer your queries.
(a.)Given $G$ is connected. We delete edges one by one from the cycles, such that graph becomes acyclic. This process must end. Note that Graph remains connected as edges which belong to the cycle are not cut-edges. We get a Spanning tree of the graph finally.
(b.)This question has 2 parts. I will prove it one by one.
Part 1: We know that every edge of a tree is a cut-edge. We delete the edge $e_1$, which then creates 2 components of $V$. Now $T_2$ is an another tree which must have some edge $e_2$ between these 2 components. So, $T_1-e_1+e_2$ is a spanning tree.
Part 2: Here, we have to show that $T_2+e_1-e_2$ is a spanning subtree. Adding an edge $e_1$ in $T_2$ creates a cycle in $T_2$. Now, we delete the some edge in that cycle which is $e_2$. Thus, $T_2+e_1-e_2$ is connected and acyclic, therefore is a spanning subtree.
Q.E.D. Hope this clarifies.