Your inquality is certainly true for $n \leq 2$. For bigger trees, consider the following lemma:
$$\#\text{leaves} = 2 + \sum_{\deg(v)\; \geq\; 2}\Big(\deg(v)-2\Big).\tag{$\spadesuit$}$$
As all the entries in the sum above are non-negative, so we obtain
$$\#\text{leaves} \geq 2 + \max_{\deg(v)\; \geq\; 2}\Big(\deg(v)-2\Big) = \max_{v \in V} \deg(v)$$
which is exactly what we need.
Proof of the $\spadesuit$ equality:
We know that the sum of all degrees is twice the number of edges and the number of edges in any tree is the number of vertices minus one, so
$$\sum_{v\in V}\deg(v) = 2m = 2(n-1).$$
Naturally $\sum_{v \in V}1 = n \cdot 1 = n$, hence
$$\sum_{v \in V}\Big(\deg(v)-2\Big)= -2.$$
Splitting $V$ into leaves and non-leaves we get
$$\sum_{\deg(v) = 1}\Big(\deg(v) - 2\Big) + \sum_{\deg(v) \geq 2}\Big(\deg(v) - 2\Big) + 2 = 0,$$
that is,
$$\sum_{\deg(v) = 1}\Big(-1\Big) + \sum_{\deg(v) \geq 2}\Big(\deg(v) - 2\Big) + 2 = 0,$$
simplifying
$$\sum_{\deg(v) \geq 2}\Big(\deg(v) - 2\Big) + 2 = \sum_{\deg(v) = 1}\Big(1\Big) = \#\text{leaves}.$$
$$\\\tag*{$\square$}$$
I hope this helps $\ddot\smile$
Your proof looks good. It's not the only way of proving this (as usual) - I would perhaps find the option to split on the root node a more natural approach for a binary tree.
I don't think induction on $N$ would be easy to frame or justify. Certainly when you're trying to prove something in which the given fact is about $L$ and the result is about $N$ you would have to do some work to turn it round.
Best Answer
A tree on $n$ vertices has $n-1$ edges. So if there are $a$ leaves and $b$ vertices of degree at least three, then by the handshaking lemma, $$a+2(n-a-b)+3b\leq\sum \deg v=2(n-1)\implies a\geq b+2.$$