The center of a graph $G$ is the vertices $v$ for which $\max_{u \in V(G)}d(v,u)$ is minimal ($d(v,u)$ is the distance from $v$ to $u$). As the title implies the I want to show that if $G$ is a tree, the set of center vertices consists of either a single vertex or two adjacent vertices. Also note that $ecc(v)$ is the eccentricity of $v$ which is the maximum distance from $v$ to any other vertex
So here is my idea for the proof. I think it is pretty rigorous, but it sort of seems like there's something missing, or like I didn't cover a special case or something:
Suppose this is not the case, that is, there exist two non-adjacent vertices in the set of center vertices (since $G$ is a tree, there will be two non-adjacent vertices if there are any more than two vertices in the set). Call these vertices $u$ and $v$ and consider the respective vertices $u'$ and $v'$ where $ecc(u) = d(u, u')$ and $ecc(v) = d(v, v')$. Recall the fact that a path from any vertex $x$ to another $y$ on a tree is unique and will be denoted here as $P_{xy}$. Observe that if $P_{uu'}$ does not go through $v$, then $P_{vu'} = P_{vu} + P_{uu'}$, which will be of larger length than that of $P_{uu'}$ (`+' here simply represents path concatenation), so $d(v,u') > \max_{t \in V(G)} d(u,u')$, implying that $v$ is not a center of $G$. Therefore $P_{uu'}$ must go through $v$. Likewise, without losing generality, $P_{vv'}$ must go through $u$. Notice that this is the case for every vertex that is "furthest" from $u$ and "furthest" from $v$ respectively since we used a general $u'$ and $v'$. Now since $uv \notin E(G)$ there exists a vertex, say $w$, in between $u$ and $v$, i.e. $w \in P_{uv}$. Observe that because $w$ is between $u$ and $v$, it follows that $d(w,u') < d(u, u')$ and $d(w,v') < d(v, v')$. So because of the strict inequalities, it follows that neither $u$ nor $v$ are centers of $G$. Hence, the set of center vertices of a tree graph must be either a single vertex or two adjacent vertices.
Thoughts? Thank you for any input.