Proof
(related to Lemma: Lower Bound of Leaves in a Tree)
 By hypothesis, \(T(V,E,\gamma)\) is a tree of order \(T\ge 2\).
 According to equivalent definitions of trees, for every pair of vertices \(u,v\in V\) there is exactly one path \(P^k:=x_0x_1\ldots x_{k1}x_k\) in \(T\) with \(x_0=u\) and \(x_k=v\).
 Choose \(u,v\) such that \(P^k\) will have a maximal length \(k\).
 Since \(P^k\) is maximal, \(\delta_T(u)\in P^k\) with \(\delta_G(u)=\{e\in E: u\in \gamma(e)\}\).
 Therefore, the degree \(d_T(u)=\delta_T(u)=1\).
 Similarly for \(v\in V\).
 Therefore, \(T\) has at least the two leaves \(u\) and \(v\).
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References
Bibliography
 Aldous Joan M., Wilson Robin J.: "Graphs and Applications  An Introductory Approach", Springer, 2000