Is this proof vigorous enough? (Feel free to provide any suggestions for improvement as well).
(A centre of a connected graph $G$ is a vertex $v$ with the property that the maximum of the distances between $v$ and all other vertices of $G$ is as small as possible.)
Note: I will not prove the tree properties listed here.
Declarations:
Let $T$ be any tree, $C$ be the set of centres of $T$, and $X$ be the set of end-vertices of $T$.
Let $a(x)$ be the vertex adjacent to $x\in X$.
Let $L(v)$ be the sum of distances between vertex $v\in T$ and all other vertices in $T$ (It is unnecessary to distinguish distance and maximum distance in a tree since there is only one path between any pair of vertices). Note that for all $c\in C$, $L(c)=\text{min}[L(v):v\in T]$.
Let $T'$ be a tree after $X$ is removed from $T$ (since no edges are added to $T$, $T'$ is still a tree) and $C'$ be the set of centres of $T'$.
Let $L'(v')$ be the sum of distances between vertex $v'$ and all other vertices in $T'$. Note that for all $c'\in C'$, $L'(c')=\min[L'(v):v'\in T']$.
Claim 1: $|X|>0$ if $|T|>1$.
Proof 1:
Assume $|X|=0.$ So, $\sum_{v\in T}\deg(v)\ge 2|T|$. By the handshaking lemma, $|E(T)|\ge |T|$. However, one of the tree properties states that $|E(T)| = |T| - 1 \le |T|$. Proven by contradiction.
Claim 2: For all $u,v\in T'$, if $L'(u)<L'(v)$, then $L(u)<L(v)$.
Proof 2:
Let $l'(a,b)$ be the distance between vertex $a$ and vertex $b$ in $T'$. So, for all $u'\in T', L'(u')=\sum_{v'\in T'}l(u',v')$.
Note that for all $u'\in T',L(u')=L'(u')+ \sum_{x\in X} [l'(u',a(x))+1]$. (The distance between $a(x)$ and $x$ is $1$).
Since $L'(u)<L'(v)\implies \sum_{x\in X} [l'(u,a(x))+1] < \sum_{x\in X}[l'(v,a(x))+1]$, $L(u)<L(v)$.
Claim 3: $C=C'$.
Proof 3:
Assume $C\neq C'$. So, for all $c\in C$ and $c'\in C'$, $L(c)<L(c')$ and $L(c)>L(c')$. This is contradictory to Claim 2.
Also, for all $x\in X$, $L(x)=L(a(x))+|T|-1$ ($x$ is always an additional distance away from all other vertices when compared to $a(x)$). $L(c)<L(a(x))<L(x)\implies L(c)<L(x)$ so $x$ cannot be a centre (it is visually obvious as well).
So $C=C'$.
Proof:
Let us keep removing end-vertices from $T$ ($T\to T'\to T'' \to ...$) as long as there are non-end vertices. So, for any tree, it will be until $|T|\le2$ (Claim 1). Hence, there will either be a single vertex or two adjacent vertices left (we cannot remove the remaining vertices, even if they are end-vertices, since it will left us with nothing)
Given that the centres of a tree are preserved after the removal of its end vertices (Claim 3) and all trees are left with one or two adjacent vertices after repeated end-vertices removal, all trees have one or two adjacent centres.