TITLE: For all $1 \leq i < j \leq k$, the subtrees $T_i$ and $T_j$ have a vertex in common. Show that $T$ has a vertex which is in all of the $T_i$.
QUESTION [6 upvotes]: Can someone please verify my proof or offer suggestions for improvement? I am aware that there is a similar question elsewhere, but I want help with my proof in particular.

Let $T_1, \ldots, T_k$ be subtrees of a tree $T$ such that for all $1 \leq i < j \leq k$, the trees $T_i$ and $T_j$ have a vertex in common. Show that $T$ has a vertex which is in all of the $T_i$.

Suppose, for the sake of contradiction, that for every vertex $v \in V(T)$, there is a subtree $T'$ in the list $T_1, T_2, \ldots, T_k$ such that $v \notin V(T')$.
Pick $v_1 \in T_1$. There is a subtree $T^{(1)}$ such that $v_1 \notin T^{(1)}$. $T_1$ and $T^{(1)}$ have a vertex $y_1$ in common. Now, there exists a tree $T^{(2)}$ such that $y_1 \notin T^{(2)}$. But now, $T^{(2)}$ and $T^{(1)}$ have a vertex $y_2$ in common. Also, $T_1$ and $T^{(2)}$ have a vertex $y_3$ in common.
There exists a path $p_1$ in $T_1$ from $y_3$ to $y_1$. Also, there is a path $p_2$ in $T^{(1)}$ from $y_1$ to $y_2$ and a path $p_3$ in $T^{(2)}$ from $y_2$ to $y_3$. Note that the path formed by the union of $p_1$ and $p_2$ passes through $y_1$, while $p_3$ does not, since $y_1 \notin T^{(2)}$. Therefore, the union of $p_1, p_2$, and $p_3$ forms a cycle, contradicting the fact that $T$ is a tree. Therefore, it must be the case that there is a vertex $x$ in all of the $T_i$.

REPLY [2 votes]: It suffices to prove the following lemma:
$\textbf{Lemma.}$ If $T_1, T_2, T_3$ are subtrees of a tree $T$ such that any two intersection is not empty, then all $3$ subtrees have a common vertex.
$\textbf{Proof.}$ Suppose no vertex lies in all $3$ trees. Then there exist distinct vertices $w_1, v_2, v_3$ in $T_2\cap T_3, T_3\cap T_1, T_1\cap T_2$ respectively, none of which belongs to their respective third tree.
So there exist a path $P \in T_3$ from $w_1$ to $w_2$. In this (unique) path, choose two vertices $x$ and $y$ in $P$ such that $x\in T_1$ and $y\in T_2$, and any vertex between $x$ and $y$ along $P$ are not in $T_1\cup T_2$. Note that $x$ and $y$ must necessarily be distinct by assumption, but are allowed to be adjacent.
Then we can define a path $Q$ from $x$ to $y$ in $T_1\cup T_2$, by taking the shortest path from the walk $x$ to $w_3$ to $y$ in $T_1\cup T_2$. This path $Q$ is disjoint from $P$ so $P\cup Q$ forms a cycle, a contradiction with $T$ being acyclic. $\square$
The problem now follows by induction: Suppose this is true for some $k\ge 3$, with base case $k=3$ proven above. Given $k+1$ subtrees with the given conditions, notice that for any $1\le i < j \le k$, $T_i\cap T_j$ is also a subtree of $T$. Indeed, if $x$ and $y$ are vertices in $T_i\cap T_j$, then take the unique path $P$ in $T$ from $x$ to $y$, $P$ must be in both $T_i$ and $T_j$ for it to be subtrees of $T$.
Now, note that by the lemma, $T_i\cap T_{k+1}$ and $T_j\cap T_{k+1}$ have a common vertex for all $i\neq j$, and are all subtrees of $T$. So we can apply the hypothesis on these subtrees to conclude they have a common vertex, hence so as the subtrees $T_1,\cdots T_{k+1}$. $\blacksquare$