Proving f'(0) has f as constant function

WTS: $\forall x\in (a,b),f^{\prime }(x)=0⟹\forall x_1,x_2\in (a,b),f(x_1)=f(x_2)$

Proof

1. Let $a,b$ be arbitrary
2. Let $a<b$
3. Assume $\forall x\in (a,b),f^{\prime }(x)=0$
4. This means $f$ is Differentiable on $(a,b)$
5. So $f$ is continuous on $(a,b)$
6. Let $x_1,x_2\in (a,b)$ be arbitrary
7. Case 1: $x_1=x_2$
   1. Then $f(x_1)=f(x_2)$
8. Case 2: $x_1<x_2$ or $x_2<x_1$
   1. Note that $f$ is Differentiable on $(x_1,x_2)$ as $(x_1,x_2)\subset (a,b)$
   2. Note that $f$ is Continuous on $[x_1,x_2]$ as $(x_1,x_2)\subset (a,b)$
   3. Thus by MVT, $f^{\prime }(x)=\frac{f(x_2)-f(x_1)}{x_2-x_1}$
   4. $⟹0=\frac{f(x_2)-f(x_1)}{x_2-x_1}$
   5. $⟹0=f(x_2)-f(x_1)$
   6. $⟹f(x_1)=f(x_2)$
9. Hence $f(x_2)=f(x_1)$
10. As required $\square$