Induction over natural numbers is a standard tool in Mathematics. But what about doing induction over real numbers´? The structure is different and it's not immediately clear what is meant, so let me clarify it.

Let $P(v)$ be a proposition that we want to show for all $v \in \mathbb{R}^+_0$.

Then we can use the following "induction principle" to prove it:

$\exists \epsilon \in \mathbb{R}^+$
$\text{a)} \forall u \in \left [0,\epsilon \right ]: P(u)$
$\text{b)} \left( \forall u \in \left [0, v - \epsilon \right ]: P(u) \right ) \Rightarrow P(v)$´

It's easy to show that this is a correct way to prove it. I'll use induction over natural numbers to prove it

Let $I_k := \left [k \cdot \frac \epsilon 2, (k+1) \cdot \frac \epsilon 2 \right ]$.
Then let's show by induction over $k$ that $\text{(*)} \forall u \in I_k: P(u)$:

Base:

From a) it follows that (*) already holds for $I_0$ and $I_1$.

Induction Step ($k \ge 2$):

If (*) holds for $I_0, ..., I_{k-1}$, then it obviously holds for all $u \in \left [ 0, k \cdot \frac \epsilon 2 \right ]$.

Fix $v \in I_k$, then $v \le (k+1) \cdot \frac \epsilon 2 \Leftrightarrow v - \epsilon \le (k-2) \cdot \frac \epsilon 2 < k \cdot \frac \epsilon 2 \right$. With b) it follows that $P(v)$ holds.

That is, $P(v)$ $\forall v \in I_k$.

#

It should be possible to show that you can generalize this to:

$\exists \epsilon \in \mathbb{R}^+$
$\text{a)} \forall u \in \left [0,\epsilon \right ]: P(u)$
$\text{b)} \left( \forall u \in \left [\psi(v), \phi( v ) \right ]: P(u) \right ) \Rightarrow P(v)$
with $\phi( v ): \mathbb{R}^+ \to \mathbb{R}^+_0$, $\psi( v ): \mathbb{R}^+ \to \mathbb{R}^+_0$ and $\psi(v) \le \phi(v) < v$.