I want prove that the problem
$$y'=f(x,y)$$ $$y(x_0)=y_0$$
It has a unique solution if $f$ is continuous in $[a,b] \times \mathbb{R}^n$ and it satisfies the Lipschitz condition with respect to the second variable. Applying the Runge-Kutta formula of order 4
My attempt
Finding the solution of the problem is equivalent to finding the solution of
$$y(x)=y_0+\int_{x_0}^{x}f(t,y(t))dt$$
By applying the Runge-Kutta method of order 4, we obtain the sequence $\left\{y_n\right\}$ given by
$$y_n=y_{n-1}+\dfrac{\Delta x}{6}\left(k_1+2k_2+2k_3+k_4\right)$$
where
$$k_1=f\left(x_0+\Delta x (n-1),y_{n-1}\right)$$ $$k_2=f\left(x_0+\Delta x (n-1) +\dfrac{\Delta x}{2},y_{n-1}+\dfrac{k_1 \Delta x}{2}\right) $$ $$k_3=f\left(x_0+\Delta x (n-1) +\dfrac{\Delta x}{2},y_{n-1}+\dfrac{k_2 \Delta x}{2}\right) $$ $$k_4=f\left(x_0 + n \Delta x, y_{n-1}+k_3 \Delta x\right)$$
It's easy to see that
$$y_n=y_0+ \sum_{m=1}^n \left(y_m-y_{m-1}\right)$$
The sequence $\left\{y_n\right\}$ is convergent in $\mathbb{R}$ if and only if the series
$$\sum_{m=1}^{\infty}\left(y_m-y_{m-1}\right)$$
is convergent.
Let's prove that the series is uniformly convergent in $[a,b]$ for this we will apply the Weierstrass test $M$, for this we must prove that there exist $M_m>0$ such that
$$|y_m-y_{m-1}| \leq M_m \quad \sum_{m=1}^{\infty}M_m < \infty$$
Let's first find $M_1$ and $M_2$ to be able to establish a recursion for the $M_m$ this is
$$|y_1-y_0| \leq M_1 \quad |y_2-y_1| \leq M_2$$
However
$$y_1=y_{0}+\dfrac{\Delta x}{6}\left(k_{11}+2k_{21}+2k_{31}+k_{41}\right)$$
where
$$k_{11}=f\left(x_0,y_{0}\right)$$ $$k_{21}=f\left(x_0 +\dfrac{\Delta x}{2},y_{0}+\dfrac{k_{11} \Delta x}{2}\right) $$ $$k_{31}=f\left(x_0 +\dfrac{\Delta x}{2},y_{0}+\dfrac{k_{21} \Delta x}{2}\right) $$ $$k_{41}=f\left(x_0 + \Delta x, y_{0}+k_{31} \Delta x\right)$$
and
$$y_2=y_{1}+\dfrac{\Delta x}{6}\left(k_{12}+2k_{22}+2k_{32}+k_{42}\right)$$
where
$$k_{12}=f\left(x_0+\Delta x ,y_{1}\right)$$ $$k_{22}=f\left(x_0 +\dfrac{ 3 \Delta x}{2},y_{1}+\dfrac{k_{12} \Delta x}{2}\right) $$ $$k_{32}=f\left(x_0 +\dfrac{3 \Delta x}{2},y_{1}+\dfrac{k_{22} \Delta x}{2}\right) $$ $$k_{41}=f\left(x_0 + 2 \Delta x, y_{1}+k_{32} \Delta x\right)$$
Therefore
$$|y_1-y_0| =\left| \dfrac{\Delta x}{6}\left(k_{11}+2k_{21}+2k_{31}+k_{41}\right) \right|$$
and
$$|y_2-y_1| = \left|\dfrac{\Delta x}{6}\left(k_{12}+2k_{22}+2k_{32}+k_{42}\right) \right|$$
Is my procedure correct? How can I continue?