Prove that if $f'(x)≠0$ on an interval $( a , b)$ on $A$ for a differentiable function $f$, then $f$ is one-to-one on $A$. Prove that if $f'(x)≠0$ on an interval $( a , b)$ on $A$ for a differentiable function $f$, then $f$ is one-to-one on $A$.

I was wondering whether the proof in the attached picture is correct. Could anyone please help me?
Not quite. If $f$ is not one-to-one on $(a,b)$ then there exist two distinct points $x,y \in (a,b)$ satisfying $f(x) = f(y)$. But there exists a point $z$ in between $x$ and $y$ satisfying $$0 = \frac{f(y) - f(x)}{y-x} = f'(z) \not= 0$$ which is the contradiction you need.

There is no reason for $f(a) = f(b)$ to hold.
Your pictured proof just shows that $f$ is injective.  You now have to show it is surjective as well.