To prove that it's false you just need a counter-example.
The two MST can be
Noticed that the only two common edge is not the lowest cost edge. Hence the statement is FALSE.
However, if all the edges have unique weights, the answer is TRUE.
Any MST of a size larger than 1 contains the smallest and the second
The smallest edge must be in any MST. Suppose not, we can always include it and form a cycle, and remove the largest edge in the cycle to form a smaller tree.
The second smallest edge must also be in any MST (of the size larger than 1). Suppose not; we can include it and form a cycle. Noticed that a cycle must have at least 3 edges. Hence, the largest edge that is removed must be larger than the second smallest edge. Hence we also form a smaller tree.
Since any MST must contain the 2 smallest edge, when they only have 2 edges in common, it must be both of them, which are also the lowest costs edge in the graph.