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.

Your proof is finished with the the line that begins "With $\;f'(c)>0\;$ ..."! Nevertheless, I would write it down as follows:
Take any two different points $\;x_1,\,x_2\in [a,b]\,,\,\,x_1<x_2\;$ ( of course, it can be $\;x_1=a\;$ or $\;x_2=b\;$). Since the conditions of the MVT are fulfilled in $\;[x_1,x_2]\;$ , there exists 
$$\;c\in (x_1,x_2)\;\; s.t. \;\;\frac{f(x_2)-f(x_1)}{x_2-x_1}=f'(c)\stackrel{\text{given!}}>0\;$$ 
Since $\;x_2-x_1>0\;$, we get that it must be $\;f(x_2)-f(x_1)>0\implies f\;$ is mon. ascending.$\;$ Q.E.D
