How to find some $\delta>0$ such that $f(x)<f(x_0)<f(y)$ whenever $x_0-\delta<x<x_0<y<x_0+\delta$ if $f^\prime(x_0)>0$ for some point $x_0$ in the interior of the domain of $f$?
I found a similar question here, but I am still confused how to find $\delta$.
My attempt:
Proof
Assume not. That is, assume $\forall\delta>0, f(x)\nless f(x_0)\nless f(y)$ whenever $x_0-\delta<x<x_0<y<x_0+\delta$ if $f^\prime(x_0)>0$ for some point $x\in$dom($f$). Then $f$ must be nonincreasing. So the $f^\prime(x_0)\leq 0$. But $f^\prime(x_0)>0$, so we have a contradiction.... but this is obviously incomplete and perhaps even wrong.
How do I find a $\delta>0$ where all the conditions are satisfied? Also, does this mean $f$ is increasing in the interval $(x_0-\delta, x_0+\delta)$? I say no because the function can switch directions at $x$ and $y$ so that $x_0-\delta>x_0+\delta$.
By the definitions of derivative and limit, there is a $\delta > 0$ such that $|x - x_0|<\delta$ implies $\frac{f(x) - f(x_0)}{x - x_0} > 0$ since $f'(x_0)>0$. Now treat the cases when $x<x_0$ and $x>x_0$.