Proofs involving subtrees of a tree I have found some claims about trees in my graph theory text, and I am wondering if corresponding proofs can be found, as I cannot find any online or in another text.
First,

If $T_1$ and $T_2$ are subtrees of tree $T$ such that $V_{T_1} \cap V_{T_2} \neq \emptyset$, then $T_1 \cap T_2$ is a subtree of $T$.

Further, this appears to be an expansion of the above claim:

If $T_1, T_2, T_3$ are subtrees of $T$ with vertex sets $V_1, V_2, V_3$ such that $V_i \cap V_j \neq \emptyset$ for each $i, j$, then $V_1 \cap V_2 \cap V_3 \neq \emptyset$ and $T_1 \cap T_2 \cap T_3$ is a subtree of $T$.

 A: A tree is a graph that is connected and does not contain any cycles. Since $T_1 \cap T_2$ is a subgraph of both $T_1$ and $T_2$, which are trees, it can obviously not contain cycles.
What remains to be shown is that $T_1\cap T_2$ is still connected. Fix two vertices $v_1, v_2$ in $T_1\cap T_2$. Obviously, we then also have $v_1, v_2\in T_1$ and $v_1, v_2\in T_2$. Furthermore, we assumed $V_{T_1}\cap V_{T_2}\neq \emptyset$. Let therefore $v$ be a vertex in $V_{T_1}\cap V_{T_2}$ (and thus also in the graphs $T_1$, $T_2$, and $T_1\cap T_2$).
Since $v_1$ and $v$ (resp. $v_2$ and $v$) are in $T_1$ (resp. $T_2$) and $T_1$ (resp. $T_2$) are connected, there is a path from $v_1$ to $v$ (resp. from $v_2$ to $v$). By combining these paths, we also have a path from $v_1$ to $v_2$.
Therefore, $T_1\cap T_2$ is connected, and because of its cycle-freeness, it is then also a tree.


For the second claim, consider this:
If we have three trees $T_1,T_2,T_3$ that are subtrees of a tree $T$ no two of which are disjoint, we can obtain elements $t_{12}, t_{13}, t_{23}$ such that $t_{ij} \in T_i\cap T_j$. Since $t_{12},t_{13}\in T_1$, $t_{12},t_{23}\in T_2$, and $t_{13},t_{23}\in T_3$, there exist paths from $t_{12}$ to $t_{13}$ within $T_1$, from $t_{12}$ to $t_{23}$ within $T_2$, and from $t_{13}$ to $t_{23}$ within $T_3$.
Since trees are free of cycles, the path from $t_{12}$ to $t_{13}$ within $T_1$ must then be equal to that from $t_{12}$ over $t_{23}$ to $t_{13}$ and thus $t_{23}\in T_1$, and, of course, also $t_{23}\in T_2\cap T_3$ and thus the $T_1\cap T_2\cap T_3 \neq \emptyset$.
Now applying the lemma for two subtrees twice yields that $T_1\cap T_2\cap T_3$ is a tree; for a more direct approach, $t_{23}$ can play the role of the node that all other nodes must be connected to, similarly to $v$ above.
EDIT: I previously thought the second claim was false, giving an incorrect counterexample.
A: The "expansion of the above claim" is a particular case of Exercise 7 on my Spring 2017 Math 5707 midterm 2:

Let $G = \left(V, E, \phi\right)$ be a multigraph.
For any subset $U$ of $V$, we let $G \left[U\right]$ denote the
  sub-multigraph $\left(U, E_U, \phi\mid_{E_U}\right)$ of $G$, where
  $E_U$ is the subset $\left\{ e \in E \mid \phi \left(e\right) \subseteq U \right\}$ of
  $E$.
  (Thus, $G \left[U\right]$ is the sub-multigraph obtained from $G$ by removing
  all vertices that don't belong to $U$, and subsequently removing all
  edges that don't have both their endpoints in $U$.)
  This sub-multigraph $G \left[U\right]$ is called the induced
  sub-multigraph of $G$ on the subset $U$.
Let $A$, $B$ and $C$ be three subsets of $V$ such that the
  sub-multigraphs $G \left[A\right]$, $G \left[B\right]$ and $G \left[C\right]$ are
  connected.
A cycle of $G$ will be called eclectic if it contains at
  least one edge of $G \left[A\right]$, at least one edge of $G \left[B\right]$ and
  at least one edge of $G \left[C\right]$ (although these three edges are not
  required to be distinct).
(a) If the sets $B \cap C$, $C \cap A$ and $A \cap B$ are
  nonempty, but $A \cap B \cap C$ is empty, then prove that $G$ has an
  eclectic cycle.
(b) If the sub-multigraphs
  $G \left[B \cap C\right]$, $G \left[C \cap A\right]$
  and $G \left[A \cap B\right]$ are connected, but the sub-multigraph
  $G \left[A \cap B \cap C\right]$ is not connected, then prove that $G$ has
  an eclectic cycle.
[Note: Keep in mind that the multigraph with $0$ vertices
  does not count as connected.]

(Why this exercise actually generalizes the claim is explained after the solution of the exercise. I am not giving a theorem number, since that number is going to shift.)
