Let $V_2=V\setminus V_1$, the set of interior vertices of the tree, and to simplify notation write $\Delta$ for $\Delta(V)$. The inequality can be rearranged to yield
$(\Delta-2)|V|+2\ge(\Delta-1)|V_1|=(\Delta-2)|V_1|+|V_1|$ and then
$|V_1|\le(\Delta-2)(|V|-|V_1|)+2=(\Delta-2)|V_2|+2\;.\tag{1}$
I’ll prove $(1)$ by induction on $|V|$.
Pick two leaves, $v_0$ and $v_n$, and let $v_0v_1\dots v_n$ be the unique path between them, which I’ll call the spine of the tree. Each of the vertices $v_1,\dots,v_{n-1}$ has degree at most $\Delta$, so it’s an end of at most $\Delta-2$ edges that aren’t in the spine. For each non-spine edge $e$ incident at $v_k$ let $T_k(e)$ be the subtree whose vertices are $v_k$ and all vertices of $T$ whose unique path to $v_k$ included the edge $e$. Let the vertex set of $T_k(e)$ be $V_k(e)$, and let $V_{k,1}(e)$ and $V_{k,2}(e)$ be the sets of leaves and interior vertices, respectively. Finally, let $t_k=\deg(v_k)-2$ be the number of subtrees incident at $v_k$; clearly $t_k\le\Delta-2$.
By the induction hypothesis we have
$|V_{k,1}(e)|\le(\Delta-2)|V_{k,2}(e)|+2$
for each of these subtrees.
Each $v_k$ is a leaf of each $T_k(e)$, so
$\begin{align*} |V_1|-2+\sum_kt_k&=\sum_{k,e}|V_{k,1}|\\ &\le\sum_{k,e}\left((\Delta-2)|V_{k,2}(e)|+2\right)\\ &=(\Delta-2)\sum_{k,e}|V_{k,2}(e)|+\sum_{k,e}2\\ &=(\Delta-2)\Big(|V_2|-(n-1)\Big)+2\sum_kt_k\;, \end{align*}$
and therefore
$\begin{align*} |V_1|&\le 2+(\Delta-2)|V_2|-(n-1)(\Delta-2)+\sum_kt_k\\ &\le 2+(\Delta-2)|V_2|\;, \end{align*}$
since $\sum_kt_k\le(n-1)(\Delta-2)\;.$