How to draw a graph to disprove this statement?
You have made a mistake interpreting the wording of the question compared to the Stackoverflow question (link in case of later editing).

Notice that in options (a) and (b), the phrasing is a minimum weight edge, whereas in the Stackoverflow question the phrasing is the minimum weight edge. In your question, there's no guarantee that the edge is the only edge with that weight, in the Stackoverflow question, there is exactly one edge with the lowest weight.

Thus the mistake is assuming that (b) is true, when in fact it is false. The two true options are (a) and (c), which is why you can't find a counter example to (c).

In both (a) and (c) however, the proof is very similar to that in the Stackoverflow question. In fact (c) is given by applying that proof inductively. (a) is given by basically the same argument, but you show that the tree made by adding the given edge and removing some other edge from the cycle can't have higher weight (because the new edge has minimum weight).

To see that (b) is false, just take the graph $K_{n}$ where all edge weights are the same. Any spanning tree is a minimum spanning tree.

For (d), consider the following graph; take a tree on $n$ vertices, give all these edges weight $1$. Then add all non-edges, give these new edges weight $2$.  Clearly there's only one MST, but there's $\binom{n}{2} - n + 1$ edges with the same weight.