$f'(x)>0$ except for countable points implies $f$ strictly increasing 
Let $f\in C[a,b]$, and except $\{x_n\}\subset [a,b]$, $f'(x)>0$. Show that $f$ is strictly increasing.

Note that we do not know the differentiability of $f$ at each $x_n$. What we know is just: $f'(x)>0, \forall\ x\not\in \{x_n\}$.
If ``except $\{x_1,\cdots,x_n\}\subset [a,b]$, $f'(x)>0$, then it is OK, but dividing the interve into finite number of small intevels, on each part $f$ is strictly increasing.
The trouble is $\{x_n\}$, a sequence...Any ideas?
 A: We first show that $f(a)\leq f(b)$. Suppose that $f(a) > f(b)$ and select $c$ such that $f(a)>c>f(b)$ and $f(x_n)\neq c$ for all $n\in \mathbb{N}$. Consider the set $$A=\{x\in [a,b] : f(x) > c\}$$ and let $\beta = \sup A$. Now if we take $y_n \in A$ with $y_n \to \beta$ and pass to the limit we get $f(\beta)\geq c$. But if $f(\beta)>c$ for sufficiently small $\epsilon > 0$ we would have $f(\beta + \epsilon)>c$, a contradiction. So we have established that $f(\beta)=c$. Now look at
$$f'(\beta) = \lim_{x\to \beta}\frac{f(x)-f(\beta)}{x-\beta}$$
For $x>\beta$ we have $f(x)\leq c$ and $f(\beta)=c$ so it must be the case  that $f'(\beta) \leq 0$ (we know that $f'(\beta)$ exists because we required that $f(x_n)\neq c$ for all $n\in \mathbb{N}$). This is a contradiction since we have $f'>0$ everywhere except possibly the $x_n$'s. So $f(a)\leq f(b)$. Now for all $x\in (a,b)$ we have $f(a)\leq f(x) \leq f(b)$ by repeating the above argument. If $f(a)=f(b)$ then $f$ would be constant and $f'=0$ so $f(a)<f(b)$.
Now for arbitrary $x,y\in [a,b]$ with $x<y$ we can repeat our argument in $[x,y]$ to get $f(x)<f(y)$, thus proving that $f$ is strictly increasing.
A: Comment only.
The idea in the proof that Giorgos has given is worth pondering over.  If there were no points at which $f'(x)$ failed to be positive, then (using the intermediate value property of continuous functions)  select any  $c\in (a,b)$ with $f(a)>c>f(b)$ and use his "last point argument" to consider
$$ \beta =\sup\{ x\in (a,b):  f(x)>c\}$$
where we are sure that $f'(\beta)>0$  etc.
If there is a set of points $E\subset (a,b)$ where we perhaps do not have a positive derivative, well... just avoid them.  If $E$ is countable, so is $f(E)$ and there is plenty of room to choose a value $c\not\in f(E)$.  That will get us a $\beta\not\in E$  etc.
That method leads to an interesting and simple generalization using the same method, due to Antoni Zygumund.  As a follow up to this "exercise" I recommend proving this theorem.

Theorem (Zygmund)  Let $f$ be a continuous function such that the set of values assumed by $f(x)$ at the points $x$ where the upper
right Dini derivative $D^+f(x)\leq 0$ contains no nondegenerate
interval.  Then $f$ is monotone nondecreasing.

