Let $P_1 = \langle v_0, \dots, v_m \rangle$ and $P_2 = \langle u_0, \dots, u_n \rangle$, such that $P_1$ and $P_2$ have no common vertices, be two different longest inextensible paths in a graph. Let us assume that $m \ge n$. Furthermore, let $v_k$ and $u_l$ be connected (in the original graph) be a path $\langle v_k, w_1, \dots, w_p, u_l \rangle$ such that neither of $w_i$ belong to either $P_1$ or $P_2$ (the length of such path is, obviously, $p+1$).
Without the loss of generality, we can assume that $k \ge m/2$ and $l \ge n/2$. This is just a matter of notation: we're labeling vertices in a way that $\langle v_0, \dots, v_k \rangle$ is longer than or equally long as $\langle v_k, \dots, v_m \rangle$ and that $\langle u_1, \dots, u_l \rangle$ is longer than or equally long as $\langle u_l, \dots, u_n \rangle$.
We construct a new path
$$P := \langle v_0, \dots, v_k, w_1, \dots, w_p, u_l, \dots u_0 \rangle$$
which has length
$$k + (p + 1) + l \ge m/2 + p + 1 + n/2 \ge n/2 + p + 1 + n/2 = n + p + 1 > n, $$
so $P$ is strictly longer than $P_2$ (and obviously different from $P_1$), which is a contradiction with the assumption that $P_2$ is the second longest inextensible path in the graph.
Let $T_1'$ be a minimal subtree of $T_1$ such that $V(T_1')\cap V(T_j)\ne\emptyset$ for all $j\in\{1,\dots,n\}$. If $T_1'$ has only one vertex, we're done. Assume for a contradiction that $T_1'$ has more than one vertex. Then $T_1'$ has at least two leaves; let $x$ and $y$ be two distinct leaves of $T_1'$. By the minimality of $T_1'$, there are indices $i$ and $j$ such that $V(T_i)\cap V(T_1')=\{x\}$ and $V(T_j)\cap V(T_1')=\{y\}$. Since $T_i$ and $T_j$ have a common vertex, their union $T_i\cup T_j$ is connected. Let $P$ be the path from $x$ to $y$ in $T_i\cup T_j$, and let $Q$ be the path from $x$ to $y$ in $T_1'$. Since $V(P)\cap V(Q)=\{x,y\}$ and $E(P)\cap E(Q)=\emptyset$, $P\cup Q$ is a cycle in $T$, contradicting the assumption that $T$ is a tree.
P.S. In the question and answer above, trees are assumed to be finite. In a comment SK19 suggested generalizing the result to infinite trees. SK19 pointed out that the result fails if $T_1,T_2,\dots$ is an infinite decreasing sequence of rays (one-way infinite paths) with empty intersection. This is essentially the only counterexample.
In what follows, trees may be finite or infinite. A tree is rayless if it does not contain a sequence of distinct vertices $v_n$ ($n\in\mathbb N$) with $v_n$ adjacent to $v_{n+1}$ for each $n$.
Lemma. A chain of rayless trees has a nonempty intersection. (The intersection, of course, is a tree.)
Proof. Assume for a contradiction that $\mathcal C$ is a chain of rayless trees whose intersection is empty. Choose a tree $T_0\in\mathcal C$ and a vertex $x_0\in V(T_0)$. Having chosen a tree $T_n\in\mathcal C$ and a vertex $x_n\in V(T_n)$, next choose a tree $T_{n+1}\in\mathcal C$ not containing $x_n$ (so $T_{n+1}$ is a subtree of $T_n$ since $\mathcal C$ is a chain), and let $P_n$ be the path in $T_n$ from $x_n$ to $x_{n+1}$. As the paths $P_0,P_1,P_2,\dots$ are internally disjoint, by concatenating them we get a ray in $T_0$, contradicting the fact that $T_0$ is rayless.
Theorem. Let $(T_i:i\in I)$ be a family of subtrees of a tree $T$. The intersection $\bigcap_{i\in I}V(T_i)$ is nonempty if and only if both of the following statements hold:
(a) $V(T_i)\cap V(T_j)\ne\emptyset$ for all $i,j\in I$;
(b) there is a rayless subtree $S$ of $T$ such that $V(S)\cap V(T_i)\ne\emptyset$ for all $i\in I$.
Proof. The "only if" is clear; we prove the "if". Suppose (a) and (b) hold; we have to show that $\bigcap_{i\in I}V(T_i)\ne\emptyset$.
Let $\mathcal P$ be the set of all subtrees $S'$ of $S$ such that $V(S')\cap V(T_i)\ne\emptyset$ for all $i\in I$. Then $\mathcal P$ is nonempty because $S\in\mathcal P$. Moreover, it follows from our lemma that every chain $\mathcal C\subseteq\mathcal P$ has a lower bound in $\mathcal P$, namely its intersection. (For $i\in I$, $\{S'\cap T_i:S'\in\mathcal C\}$ is a chain of rayless trees.) Therefore, by Zorn's lemma, $\mathcal P$ has a minimal element. Let $S_0$ be a minimal element of $\mathcal P$.
If $S_0$ has only one vertex, we're done. Assume for a contradiction that $S_0$ has more than one vertex. Then, since $S_0$ is rayless, it has at least two leaves (the endpoints of any maximal path in $S_0$); let $x$ and $y$ be two distinct leaves of $S_0$. By the minimality of $S_0$, there are indices $i,j\in I$ such that $V(T_i)\cap V(S_0)=\{x\}$ and $V(T_j)\cap V(S_0)=\{y\}$. Since $T_i$ and $T_j$ have a common vertex, their union $T_i\cup T_j$ is connected. Let $P$ be the path from $x$ to $y$ in $T_i\cup T_j$, and let $Q$ be the path from $x$ to $y$ in $S_0$. Since $V(P)\cap V(Q)=\{x,y\}$ and $E(P)\cap E(Q)=\emptyset$, $P\cup Q$ is a cycle in $T$, contradicting the assumption that $T$ is a tree.
Corollary. Let $(T_i:i\in I)$ be a family of subtrees of a tree $T$ such that $V(T_i)\cap V(T_j)\ne\emptyset$ for all $i,j\in I$. If $I$ is finite, or if $T_i$ is rayless for some $i\in I$, then $\bigcap_{i\in I}V(T_i)\ne\emptyset$.
Best Answer
No, you proved that every pair of longest paths has a common vertex. You didn't prove anything about the intersection of all longest paths that may be empty. Moreover the second part is wrong for general connected graph, because two paths may have common vertices before and after their middle vertices. In this case there is no way to extend any of this path just going through the middle vertex of the other path. Consider a path graph on $7$ vertices $v_1, v_2, \ldots, v_7$ and add one more vertex $u$ adjacent to $v_3$ and $v_5$. This graph contains exactly two longest path $v_1v_2v_3v_4v_5v_6v_7$ and $v_1v_2v_3uv_5v_6v_7$. This paths share each of $6$ vertices other than the middle one.
P. S. Also there is a small inaccuracy. I guess you mean $u_i$ to be the last vertex of $Q$ belonging to $P_1$. But it could have the smallest index $i$ among all vertices of $u_i \in V(P_1) \cap V(Q)$, if $Q$ starts for example from $u_k$ and goes through $u_{k - 1}$. The same situation with $j$.