Let $T$ be a local minimum spanning tree that is not a global minimum.
Let $e_1,\ldots,e_m$ the edges of $T$ ordered by non-decreasing weight.
Let $T'$ be a minimum weight spanning tree that coincides with $T$ on the largest possible begin segment,
so we may assume $T'$ has edges $e_1,\ldots,e_{n-1},f_n,\ldots,f_m$ (again ordered by non-decreasing weight)
and $f_n\ne e_n$. Let $w$ be the weight function in our graph.
We claim that $w(f_n)<w(e_n)$: indeed, if $w(e_n)\leq w(f_n)$ then $e_1,\ldots,e_{n-1},e_n$ would
be a valid startup for Kruskal's algorithm and would lead to a minimum weight spanning tree that
coincides with $T$ on a larger initial segment.
$T+f_n$ contains exactly one cycle $C$. The edges of $C$ different from $f_n$ cannot all be in
$\{e_1,\ldots,e_{n-1}\}$ or we would have a cycle in $T'$. So we find at least one edge $f$ on $C$ that
is different from $e_1,\ldots,e_{n-1}$ and $f_n$. But then $w(f)\geq w(e_n)>w(f_n)$,
so $T+f_n-f$ is a spanning tree with a smaller total weight that is a neighbour of $T$ in the spanning tree graph. Contradiction.
(I left some small details for you to prove. Let me know if they cause trouble).
Note that this proves the slightly stronger statement, that each non-minimal tree has
a neighbour of strictly smaller total weight.
Part $(b)$ is only true if we assume that $G$ has multiple edges but no loops.
The Wikipedia page is a little misleading; it might be more accurate to say that at least one of the lowest weight edges will be in a minimum spanning tree (in a loopless graph).
This is apparent from the application of Kruskal's Algorithm for constructing a minimum spanning tree. Essentially, select the lowest weight vertex at each step that will not introduce a cycle. Because it is impossible to create a cycle when selecting the first edge in a loopless graph (per assumption) Kruskal's Algorithm will always select it.
In order to rule out the length $10$,$10$, $11$, and $13$ edges, they must all create cycles if they are included with the $8$ edge; this implies that there is a multiple edge with $8$,$10$,$10$,$11$, and $13$ all connecting the same two vertices.
I would say it's very non-standard to allow multiple-edges but not loops in the graph, because problems are generally a matter of simple vs. non-simple, where loops are allowed in non-simple graphs. Clearly, you can obtain a higher result by making $8$,$10$,$10$, and $11$ edges all loops, while using the other $4$ to create the spanning tree.
I would ask your professor to clarify this point, and definitely ask during an exam if a question is vague.
Best Answer
Proof:
Let $G$ be a graph on $n$ vertices with $\alpha(G)\geq 2$. Let $T$ be a spanning tree of $G$ with as few leaves as possible; denote the number of leaves of $T$ as $t$.
Suppose that $t>\alpha(G)$, then there exists two leaves in $T$, say $x,y$, such that $xy$ is an edge in $G$.
Consider the graph $T+xy$ obtained by adding the edge $xy$ to $T$; note that $T+xy$ has $t-2$ vertices of degree $1$. Since $T$ is a tree, ie a maximal acyclic graph, then there exists a cycle in $T+xy$ which contains $x$ and $y$, say $v_0v_1v_2...v_mv_0$, with $v_0=x$ and $v_m=y$.
Consider now the vertices $v_1,v_2,v_3...,v_{m-1}$ in $T$. If all of them have degree 2 then $xv_1v_2...v_{m-1}y$ is a connected component of $T$; since $T$ is connected this imples that $T=xv_1v_2...v_jy$ ie $T$ is a path. But a path has exactly $2$ leaves which implies $2>\alpha(G)$, contradiction.
Thus there exists some $i\in\{1,2,3,...,m-1\}$ such that $v_i$ has degree greater than $2$. Consider the graph $T'=T+xy-v_iv_{i+1}$ obtained by deleting the edge $v_iv_{i+1}$ from $T+xy$.
$T'$ is connected (since we removed an edge in a cycle) and has $n-1$ edges; thus $T'$ is a tree.
Note that by construction the degree of $v_i$ in $T'$ is grater than $1$, hence $T'$ has at most $t-1$ leaves.
We have constructed a spanning tree $T'$ with at most $t-1$ leaves. This contradicts the initial assumption that $T$ is a spanning graph with as few leaves as possible. Thus $t\leq\alpha(G)$.