I previously posted a similar question and it seems it doesn't lead to the same conclusion if $f$ is not continuous. if $f'(c)<0$ then there exists a nbd $I$ of $c$ such that $f'(x)<0$ for all $x\in I$ . I understand now why it doesn't hold.
Here's an attempt of proof with $f'$ being continuous in a nbd of $c$:
Since we have that $f'(c)>0$, there must be $\delta >0$ such that $f'(x)>0$ for all $x \in (c-\delta , c+\delta )$. This is obtained by letting $\varepsilon := f'(c)/2 > 0$ and using the definition of continuity. Since $f'(x)>0$ for all $x \in (c-\delta , c+\delta )$, it must be that $f$ is strictly increasing in this particular nbd of $c$. This completes the proof.
Is this okay? Or am I still going wrong anywhere?