Identify a vertex of greatest degree and consider sawing the tree into $\Delta$ branches from that vertex. You then have a forest of $\Delta$ smaller trees. You have introduced exactly $\Delta$ new leaves to the forest. What is an easy lower bound on the number of leaves that the entire forest must have?
Edit: Your hint seems to be an off-shoot of the above. Applying the above method once will prove that $|V_1| \ge \Delta$. But strangely, your hint actually proves something stronger (without mentioning it).
Suppose you have $|V_\Delta|$ vertices of degree $\Delta$, then splitting the branches the first time will generate $\Delta$ new leaves with a total of $\Delta -1$ new trees. A second split again introduce $\Delta$ new leaves with another $\Delta - 1$ new trees. If you split each of the |V_\Delta| vertices of degree $\Delta$ then you end up with $1 + (\Delta - 1)|V_\Delta|$ trees in your forest and a total of $\Delta|V_\Delta|$ new leaves. Each tree in your forest has at least two leaves, therefore your entire forest has at least
$$2 + 2(\Delta - 1)|V_\Delta|$$
total leaves, of which
$$2 + 2(\Delta - 1)|V_\Delta| - \Delta|V_\Delta| = 2 + (\Delta - 2)|V_\Delta|$$
are original to your tree. So your hint proves a bound not just for a single vertex of degree $\Delta$, but an arbitrary number. This bound is easily seen to be optimal.