[proofplan] The argument is the classical longest-path trick. We extract a longest path $P$ in $T$ — this exists because $T$ is finite, so path lengths are bounded. We then inspect the two endpoints of $P$. At each endpoint, the maximality of $P$ restricts the possible neighbours: any neighbour outside $P$ would extend $P$, and any neighbour inside $P$ (other than the adjacent vertex on $P$) would close a cycle, contradicting acyclicity. The only remaining possibility is the single neighbour along $P$, making each endpoint a leaf. Since $P$ has at least $2$ vertices (as $|T| \ge 2$ and $T$ is connected), its two endpoints are distinct, giving two leaves. [/proofplan]

[step:Extract a longest path in $T$] Since $T$ has $|T| \ge 2$ vertices and $T$ is connected, there exist distinct $u, v \in V(T)$ and, by connectedness together with [Walk Contains Path](/theorems/2007), a $u$–$v$ path in $T$. So the set $\mathcal{P}$ of paths in $T$ is non-empty and contains a path of length $\ge 1$. The length of any path in $T$ is bounded above by $|V(T)| - 1$ (a path has pairwise distinct vertices, so its length, the number of edges, is at most $|V(T)| - 1$). The set of path lengths is a non-empty subset of $\{0, 1, \ldots, |V(T)| - 1\}$, hence has a maximum by finiteness. Fix a path \begin{align*} P = x_1, x_2, \ldots, x_k \end{align*} of maximum length $k - 1$ in $T$, with the vertices $x_1, \ldots, x_k$ pairwise distinct and each $x_{i-1} x_i \in E(T)$. Since the maximum length is $\ge 1$, we have $k \ge 2$, so $x_1 \neq x_k$. [/step]

[step:Show $x_k$ is a leaf by ruling out all other neighbours]We prove the neighbourhood of $x_k$ is exactly $\{x_{k-1}\}$, so $\deg_T(x_k) = 1$ and $x_k$ is a leaf. Let $w \in N_T(x_k)$. We show $w = x_{k-1}$. **Case A: $w \notin \{x_1, \ldots, x_k\}$.** Then the sequence $x_1, x_2, \ldots, x_k, w$ has $k+1$ pairwise distinct vertices (the new vertex $w$ differs from all earlier $x_i$), every consecutive pair is an edge of $T$ (the edges $x_{i-1} x_i$ are in $T$ by assumption, and $x_k w \in E(T)$ since $w \in N_T(x_k)$), and it is thus a path in $T$ of length $k$. This exceeds the maximum path length $k - 1$, contradicting the choice of $P$. **Case B: $w = x_i$ for some $1 \le i \le k - 2$.** Then consider \begin{align*} C = x_i, x_{i+1}, x_{i+2}, \ldots, x_{k-1}, x_k, x_i. \end{align*} The internal vertices $x_i, x_{i+1}, \ldots, x_k$ are pairwise distinct (as part of the path $P$). The consecutive edges $x_j x_{j+1}$ for $i \le j \le k - 1$ are edges of $T$. The closing edge $x_k x_i = x_k w$ is an edge of $T$ since $w \in N_T(x_k)$. The length of $C$ is $(k - i) + 1 \ge 3$, since $i \le k - 2$ gives $k - i \ge 2$. Hence $C$ is a cycle in $T$, contradicting the acyclicity of $T$. **Case C: $w = x_{k-1}$.** This is consistent with $w$ being a neighbour of $x_k$ via the edge $x_{k-1} x_k$ already in $P$. (Note $w \neq x_k$ since $T$ is simple, so has no loops.) Cases A and B are excluded; Case C is the only possibility. So $N_T(x_k) \subseteq \{x_{k-1}\}$, and since $x_{k-1} x_k \in E(T)$, we have equality: $N_T(x_k) = \{x_{k-1}\}$. Therefore $\deg_T(x_k) = 1$, and $x_k$ is a leaf.[/step]

[guided]We inspect each vertex $w \in N_T(x_k)$ and rule out every possibility except $w = x_{k-1}$. First, a sanity check: $T$ is a simple graph by convention, so no loops, meaning $w \neq x_k$. And $w \in V(T)$, so $w$ is some vertex of $T$. We split according to where $w$ sits relative to the path $P$: **Case A: $w$ is not on $P$**, i.e., $w \notin \{x_1, x_2, \ldots, x_k\}$. Then we can **extend** $P$ by tacking $w$ onto its end. Consider the sequence \begin{align*} x_1, x_2, \ldots, x_k, w. \end{align*} The vertices are pairwise distinct: $x_1, \ldots, x_k$ are distinct because $P$ is a path, and $w$ is distinct from all of them because $w \notin \{x_1, \ldots, x_k\}$. Every consecutive pair is an edge of $T$: the pairs $x_{i-1} x_i$ for $2 \le i \le k$ because $P$ is a path in $T$, and the new pair $x_k w$ because $w \in N_T(x_k)$. Hence this sequence is a path in $T$. Its length is $k$ (number of edges). But $P$ was chosen to be a path of maximum length $k - 1 < k$. Contradiction. **Case B: $w = x_i$ for some $i \in \{1, 2, \ldots, k - 2\}$.** Then we can **close a cycle** using the segment of $P$ from $x_i$ to $x_k$ and the new edge $x_k w = x_k x_i$. Consider \begin{align*} C = x_i, x_{i+1}, \ldots, x_{k-1}, x_k, x_i. \end{align*} Let us verify this is a cycle. The internal vertex list $x_i, x_{i+1}, \ldots, x_k$ consists of distinct vertices of $P$. The consecutive pairs \begin{align*} x_i x_{i+1},\ x_{i+1} x_{i+2},\ \ldots,\ x_{k-1} x_k \end{align*} are edges of $T$ from the path $P$, and the closing pair $x_k x_i$ is the edge $x_k w$, which lies in $T$ because $w \in N_T(x_k)$. The length is $(k - i) + 1$, where $k - i \ge 2$ since $i \le k - 2$, so the length is at least $3$. Therefore $C$ is a cycle in $T$. But $T$ is acyclic. Contradiction. **Why we need $i \le k - 2$ and not merely $i \le k - 1$.** The case $i = k - 1$ would give the "closed walk" $x_{k-1}, x_k, x_{k-1}$, which is not a cycle — it has length $2$, and a cycle must have length at least $3$. So $w = x_{k-1}$ is allowed; it corresponds to the edge $x_{k-1} x_k$ already on $P$ and creates no cycle. This is why $i = k - 1$ is excluded from Case B. **Case C: $w = x_{k-1}$.** This is permitted: the edge $x_{k-1} x_k$ is already an edge of $P$, and we cannot rule it out using maximality or acyclicity. Combining the cases, the only admissible neighbour of $x_k$ is $x_{k-1}$, so $N_T(x_k) = \{x_{k-1}\}$ and $\deg_T(x_k) = 1$, i.e., $x_k$ is a leaf.[/guided]

[step:Apply the same argument at $x_1$ to obtain a second leaf] The argument of the previous step was symmetric in the two endpoints of $P$: at no point did we use the orientation of $P$ beyond the fact that $x_k$ is an endpoint. Applying the identical argument to $x_1$ (viewed as the terminal vertex of the reversed path $x_k, x_{k-1}, \ldots, x_1$, which is a path of the same maximum length) gives $N_T(x_1) = \{x_2\}$, so $\deg_T(x_1) = 1$ and $x_1$ is a leaf. Since $k \ge 2$ ensures $x_1 \neq x_k$, the vertices $x_1$ and $x_k$ are two distinct leaves of $T$. Therefore $T$ has at least two leaves. [/step]