Need explanation on graph theory problem This is the reproduction of a problem in Pablo Soberon's combinatorics book. He stated that we will consider all the graphs in the book to be simple and finite.

Example 4.2.7. Let $G$ be a connected graph where every vertex has degree greater than or equal to $2$. Show that there are two adjacent
vertices $v_1$,$v_2$ such that, if we remove them, the remaining graph
is connected.
Solution. Let $T$ be a spanning tree of $G$. Given two vertices $v_1$, $v_2$ in $G$, there is a unique path that goes from $v_1$ to
$v_2$ in $T$ (if there were two, there would be a cycle). Let $P =
> (v_1,v_2,...,v_k)$ be the longest path in $T $. Let $u_1,u_2,...,u_l$
the vertices adjacent to $v_2$ in $T$ different from $v_1$ and $v_3$.
Note that the degree in $T$ of $v_1,u_1,u_2,...,u_l$ is $1$. If that
is not true, we can construct a path in $T$ longer than $P$ ,
contradicting its maximality.
Thus, if we remove any vertices of $v_1,u_1,u_2,...,u_l$ the
connectedness of $T$ (and thus of $G$) is not broken. If any two of
those vertices are adjacent in $G$, we are done. If not, remember that
every vertex in $G$ has degree at least 2, so every vertex of
$u_1,u_2,...,u_l$ must be adjacent (in $G$) to a vertex different from
$v_1,v_2,u_1,u_2,...,u_l$. Thus, by removing $v_1$,$v_2$ we are not
breaking the connectedness of $G$.

I could understand everything until the second paragraph. Could you please help me understand this paragraph better? Thank you so much.
 A: The vertices $v_1$ and $u_1,\ldots,u_\ell$ are leaves of the spanning tree $T$, so removing any of them does not disconnect $T$ and therefore does not disconnect $G$, either: you can still get from any remaining vertex to any other remaining vertex via $T$, so you can certainly do so via $G$.
Now suppose that two of these vertices are adjacent in $G$. We’ve just seen that we can remove them without disconnecting $G$, so they are exactly what we want: adjacent vertices whose removal does not disconnect $G$. If they aren’t adjacent, we have to work a bit harder.
By hypothesis each of the vertices $u_1,\ldots,u_\ell$ has degree at least $2$, and each of them is adjacent to $v_2$. Each of them must be adjacent to at least one other vertex, a vertex that is not $v_2$. None of $u_1,\ldots,u_\ell$ can be adjacent to $v_1$, either: if some $u_i$ were adjacent to $v_1$, then $v_1,v_2$, and $u_i$ would form a cycle in the tree $T$, which is impossible. And if $1\le i<j\le\ell$, $u_i$ and $u_j$ cannot be adjacent, because then $v_2,u_i$, and $u_j$ would form a cycle in $T$.
Note that the path in $T$ from $v_3$ to any of the vertices $v_1,v_2,u_1,\ldots$, or $u_\ell$ goes through $v_2$, while the path in $T$ to any other vertex does not. Suppose that we remove the adjacent vertices $v_1$ and $v_2$. This breaks the remainder of the spanning tree $T$ into $\ell+1$ components: each of the leaves $u_1,\ldots,u_\ell$ is now an isolated vertex, and everything else that remains, which I’ll call $T'$, is still connected, since every remaining vertex except $u_1,\ldots,u_\ell$ can still be reached from $v_3$ by a path in $T'$. Finally, we just saw in the previous paragraph that each of the vertices $u_1,\ldots,u_\ell$ is adjacent in $G$ to a vertex in $T'$, so what remains of $G$ is still connected.
