Please correct me where wrong:
Assumption 1 to the solution:
Answer (b):
This can be prove as follows:
Direct Proof:
Let us consider the longest path {v0, v1},{v1, v2},...,{vn1, vn} between two vertices x = v0 and y = vn in the tree (here the length of a path is how many edges it uses, and if there are multiple longest paths then we just pick one of them).
We claim that x and y must be leaves.
Suppose the contrary that x is not a leaf, so it has degree at least two (as we know that every node except root or leaf has at least degree 2).
This means x is adjacent to another vertex z different from v1. Observe that z cannot appear in the path from x to y that we are considering, for otherwise there would be a cycle in the tree. Therefore, we can add the edge {z, x} to our path to obtain a longer path in the tree, contradicting our earlier choice of the longest path. Thus, we conclude that x is a leaf.
By the same argument, we conclude y is also a leaf.
Thus we conclude that every tree with at least two vertices has at least two leaves.
Assumption 2 to the solution:
b)
If a tree has only 2 vertices.. then their degrees must be one...based on the definition on the tree...
so..we have 2 vertices with the degree two means..we have two leaves...
so...for every tree with at least two vertices has at least two leaves
Please could someone tell me which one is better and add details that may need specifying, thanks
(Part a my overall assumption)

A graph G is connected if for every pair of vertices u and v of G there is a path
from u to v. Otherwise G is disconnected.

A tree is a connected graph with no cycles.

A leaf is a vertex of degree 1.