Proof about monotonicity of functions 
Let $f:[a,b]\rightarrow\mathbf{R}$
Suppose $f'(x)>0$ for all x $\in (a,b)$
$\ \ \ \ \ \ \ $and $f$ is differentiable on $(a,b)$
$\ \ \ \ \ \ \ $and $f$ is continuous on $[a,b]$
Show $f$ is strictly increasing on $[a,b]$.


What i tried so far:
Let $x_1,x_2 \in (a,b)$,
Assume $x_1<x_2$
By MVT exist $c$ in $(a,b)$ s.t. $f'(c) = \dfrac{f(x_2) - f(x_1)}{x_2 - x_1}$
With $f'(c)>0$ implies $f(x_1)<f(x_2)$
That $f$ is strictly in creasing on $(a,b)$
Also WTS for all x in (a,b), $f(a)<x<f(b)$
That is $f(a)<x$ and $x<f(b)$
Assume its negation is true
Have $f(a)\geq x$ or $x \geq f(b)$
Now need to find some contradiction
...
Any help would be appreciated.
Please tell me if there is an easier proof.

If $x>a$ then $f(x)-f(a)=\lim [f(x)-f(a+\frac 1 n)] \geq 0$. If $f(x) =f(a)$ then there exists $c \in [a,x]$ such that  $f'(c)=0$ but this contradicts the hypothesis. Hence $f(a) <f(x)$. Simialrly $f(x) <f(b)$ if $x <b$.
