Let $f$ be twice differentiable function on $(0,1)$.

Let $f$ be twice differentiable function on $(0,1)$. Given that for every $x\in(0,1)$, $|f''(x)|\leq M$ where $M$ is a non-negative real number. Prove that $f$ is uniformly continuous on $(0,1)$.

My Approach:

Since $f$ is twice differentiable function on $(0,1)$, it follows that $f'$ is continuous in $(0,1)$. So, by Mean Value Theorem, for any $x,y\in (0,1)$ with $x<y$, there is a $z$ with $x<z<y$ such that $|\frac{f'(x)-f'(y)}{x-y}|=|f''(z)|$. That is, $|f'(x)-f'(y)|\leq M|x-y|$ for all $x,y\in(0,1)$.

Now , I don't know how to proceed further.