Bounded function and second derivative implies bounded derivative.
Solution 1:
I think you're almost done: for any interval $[x,y]$ with $y-x=1$, you have some $\xi \in [x,y]$ such that $| f'(\xi)| = |f(y)-f(x)| \le 2$. Now, if you pick any other point $\alpha \in [x,y]$, from the second equation you get the estimate $|f'(\alpha)-f'(\xi)| < 1$, hence $|f'(\alpha)| < 3$ for all $\alpha \in [x,y]$. But $x$ and $y$ were arbitrary points with distance $1$, so that $|f'|$ is bounded by $3$ on the whole real axis.
Exercise: try to reduce the bound to $2\sqrt{2}$.