Prove that if $f$ is strictly increasing at each point of (a,b), then $f$ is strictly increasing on (a,b). If $c \in (a,b)$ we can say that $f$ is strictly increasing at c if there exists $\delta \gt 0$ such that if $c-\delta \lt x \lt c$, then $f(x) \lt f(c)$ and if $c \lt x \lt c + \delta$, then $f(c) \lt f(x)$.  
We say that $f$ is strictly increasing on (a,b) if whenever $x,y \in (a,b)$ with $x \lt y$, we have $f(x) \lt f(y)$
Prove that if $f$ is strictly increasing at each point of (a,b), then $f$ is strictly increasing on (a,b).  
Hint: Argue by contradiction. Then there exists $c,d \in (a,b)$ with $c \lt d$ and $f(c) \ge f(d)$. Now consider $$\text{lub} \{x|a\lt x \lt d \; \text{and} \; f(x) \ge f(d)$$  I know that we have an IF,THEN type of proof so I can assume it is continuous at each point, but I am unsure as how to begin proving it using delta (or by contradiction using least upper bound hint). Any help with how to begin this problem would be appreciated.
 A: I am expanding my comment into an answer. The thing you call lub is also called supremum by many authors and I will use this term. Let $$A=\{x\mid a<x<d, f(x) \geq f(d) \}, \, s=\sup A$$ and by assumption the set described above is non-empty (it contains $c$) and is bounded above by $d$ so by completeness property the supremum $s$ exists and $c\leq s\leq d$.
We next show that $s=d$. Suppose on the contrary that $s<d$. Then $f$ is increasing at $s$ and therefore there is a $\delta_s>0$ such that if $s-\delta_s<x<s$ then $f(x) <f(s) $ and if $s<x<s+\delta_s$ then $f(s) <f(x) $. And clearly we can choose $s+\delta_s<d$. Now by definition of supremum there is an element  $p\in A$ such that $s-\delta_s<p$ and hence $f(p) \geq f(d)$ (by definition of $A$) and from what we have described above we have $f(s) \geq f(p) $. Thus $f(s) \geq f(d) $ and thus for all points $x$ with $s<x<s+\delta_s$ we have $f(x) >f(s) \geq f(d) $ so that these points $x$ belong to $A$. This is a contradiction as these points are greater than $s=\sup A$. Thus $s=d$ and then $s\notin A$. 
Next we get another contradiction because we have a point $p\in A$ with $s-\delta_s<p<s$ such that $f(p) \geq f(d) $ and $f(p) <f(s) =f(d) $. And this completes our proof.
You should also try to prove this nice result using Heine Borel theorem which is especially designed to obtain global properties from local ones. 
