Let $f$ is defined on open interval $(a,b)$ and assume that for each interior point $x$ of $(a,b)$ there exist open ball on which $f$ is increasing .Then I wanted to prove that $f$ is increasing through out open interval.
So $a<x_1<x_2<b$. I have compact set [$x_1,x_2$] So there is open cover by $B(t_i,\delta_i)$ So there exist finite cover of [$x_1,x_2$] This can be given by $\bigcup_{i\in N} B_t$ where this N is finite . consider $x_1 \in$ $B(t_1,\delta_1)$
$f(x_1)<f(t)$ for t$\in B(t_1,\delta_1)$ $f(x_1)<f(t_1)$ Similarly I can show by inducation other At last$ f(x_1)<f(x_2)$ How to tackle remaining portion?
Hint. For each $x\in (a,b)$, let $B_x=(x-r_x,x+r_x)$ with $r_x>0$ be the open ball where $f$ is increasing. Let $a<s<t<b$ then $\bigcup_{x\in (a,b)} B_x$ is an open cover of the compact set $[s,t]$. Hence there exist a finite number of points $x_1<x_2<\dots<x_n$ such that $\bigcup_{i=1}^n B_{x_i}\supset [s,t]$. We may assume that the finite covering is minimal which implies that $B_{x_i}\cap B_{x_{i+1}}\not=\emptyset$ for $1\leq i<n$.
Can you show now that $f(s)\leq f(t)$?