Show that $f$ is increasing on $(a, b)$ if and only if $f^{\prime}(x) \geq0$ for all $x \in (a, b)$. A function $f : (a, b) \rightarrow \mathbb{R}$ is said to be increasing on $(a, b)$ if $f(x)\leq f(y)$ for all $x &lt; y \in (a, b)$. Suppose that $f$ is differentiable on $(a, b)$. Show that $f$ is increasing on $(a, b)$ if and only if $f^{\prime}(x) \geq0$ for all $x \in (a, b)$.

For the backward direction, I try to use Mean Value Theorem to show but I don't know how to show that $f$ is continuous on $[a,b]$. Can anyone guide me ?
You don't need continuity on $[a,b]$, you only need it on $[x,y]$. And you have that, since $x$ and $y$ are in $(a,b)$, and our function is differentiable and therefore continuous on $(a,b)$. So it is continuous on $[x,y]$. 

Now apply the MVT to conclude that $\frac{f(y)-f(x)}{y-x}=f'(c)$ for some $c$ in the interval $(x,y)$, and use the fact that $f'(c)\ge 0$.

You did not ask about the other direction, so I assume that is under control.
if a function is differentiable on (a, b), then it is continuous on (a,b). so the condition's given, you don't have to prove it. 
and for the converse part, you have to use the definition, not the mean value theorem cause mvt saying about ONE C in between x,y. the converse part you need to prove f'(x) >0 for ALL c in(a,b).