Is it a new type of induction? (Infinitesimal induction) Is this even true?

Here is an example of your logic being used to prove a false statement. I am just basically going to take your exact proof, except replace the constant $e$ with the constant $2$. The logic is all the exact same, and you should notice that no where in your proof do you ever use any properties of the number $e$, so in theory any number could take its place.

Prove (The false euler equation): $$2^{ix}=\cos x+i\sin x \ \ \ \forall \ \ x\geq0$$

For $x=0$, we have $$1=1$$ So the equality holds.
Now let us assume that the given equality holds for some $x=k$.
$$2^{ik}=\cos k+i\sin k$$ Now, this is where I added my "own" axiom. Please answer whether this "axiom" is true or not. Now this equality must hold for $x=k+\Delta k$ also, for some infinitely small positive change $\Delta k$ (infinitesimal).
So $2^{i(k+\Delta k)}=2^{ik}.2^{i\Delta k}=(\cos k+i\sin k)(\cos\Delta k+i\sin\Delta k)$
$$=\cos k\cos\Delta k-\sin k\sin\Delta k+i\sin k\cos\Delta k+i\cos k\sin \Delta k$$ $$=\cos(k+\Delta k)+i\sin(k+\Delta k)$$ So we proved it for $x=k+\Delta k$, and hence it must hold for all $x\geq0$.

But the statement that we have just "proved" is very clearly false.

An argument along these lines is the following: the standard theory of existence and uniqueness of solutions to ODEs implies that a solution to the ODE $y' = iy$ (where $y$ is a function $\mathbb{R} \to \mathbb{C}$) is uniquely determined by its initial value $y(0)$. Now, $e^{ix}$ and $\cos x + i \sin x$ are both solutions to this ODE, and they both have initial value $y(0) = 1$.

The way in which this comes to look like an inductive argument is when you explicitly solve this equation using Euler's method.

In general, there is a "principle of real induction," and it looks something like this: suppose $P(x)$ is a property of a nonnegative real number $x$ such that

• $P(0)$ is true,
• If $P(x)$ is true, then $P(y)$ is true for all $y$ in some interval $[x, x + \varepsilon)$, where $\varepsilon > 0$.
• If $P(x)$ is true for all $x < y$, then $P(y)$ is true.

Then $P(x)$ is true for all $x$. However, it's unclear to me how easy it is to apply this to your case. (Several variations of this are possible; see, for example, this note by Pete Clark.)

The first (not biggest) flaw in you proof is when you go from $k$ to some $x=k+\Delta k$. You are trying to prove some statement for ALL x. So you'd better proove it for ALL $x$ between $k$ and $x=k+\Delta k$. What if there are some $x$ in between for which the statement does not hold? For example, for each rational number there exists even bigger rational number, but still irrational numbers also exist!

To the bigger flaw now.

Let me "proove" much simplier statement: all real numbers are strictly less than 1.

Step 1: Statement holds for $x$ = 0 and all the smaller numbers.

Step 2: for any $x < 1$ there exists some positive change $dx$ such that the statement holds for all the numbers smaller than $x+dx$. F.e. $dx = (1 - x)/2$ will do.

Step 3. So that's it? For any number which satisfies our condition there exists even bigger number which also satisfies, so all the numbers must satisfy it?

Of course not, I know several numbers that are bigger than 1. (By the way, isn't it "Achilles and the tortoise" paradox?)

Moral of this story is that mathemetical induction isn't so obvious as it looks. It is a very important property of the set of integer numbers that you can use induction on it.