For any finite interval $I$, $f$ satisfies: $\forall x$, $\exists\delta>0$ such that $f((x-\delta,x+\delta)\bigcap I)$ has upper bound, prove f([a,b]) has upper bound, for any $0<b-a<\infty$
Approach: I try using Heine–Borel theorem to show $[a,b]\subset\bigcup_{i=1}^{n}U_{\alpha_i}$ where $U_{\alpha_i}$ an open subcover. Then what remain to show will be $f(U_{\alpha_i})$ has upper bound but I get stuck. Any Hint? Thanks in advance.
Hint: Heine-Borel tells us that $[a,b]$ is compact, that is every open cover has a finite subcover. For each $x \in [a,b]$ there is a $\delta_x$ such that $f((x-\delta_x, x+\delta_x)\cap [a,b])$ has an upper bound $M_x$ by assumption. Try taking a finite subcover of the open cover $\mathcal{U} = \{(x-\delta_x, x+\delta_x) : x \in [a,b] \}$ for $[a,b]$