Prove that if a function is increasing on [a,b] and satisfied the intermediate value property, then that function is continuous on [a,b]. Let $f$ be increasing on $[a,b]$, i.e. for all $x<y$ in $[a,b]$, $f(x)\leq f(y)$. Also assume $f$ satisfies the intermediate value property. Show that $f$ is continuous on $[a,b]$.
My attempt: Let $f$ be increasing on $[a,b]$ and satisfy IVP, i.e. $\forall x<y$ on $[a,b]$ and $\forall L$ between $f(x)$ and $f(y)$, $\exists c\in(x,y)$ with $f(c)=L$.
Let $\epsilon>0$. Let $c\in(a,b)$. To show $f$ continuous on $[a,b]$, we need to choose a $\delta>0$ such that whenever $|x-c|<\delta$, $|f(x)-f(c)|<\epsilon$.
Since $f$ is increasing and $a<c$, then $f(a)\leq f(c)$. I want to show that $f(c)-\epsilon\leq f(x)\leq f(c)+\epsilon$ but don't know how.

Consider a point $c\in[a,b]$ where $f$ is supposed not continous. So $\exists\epsilon_0>0$ such that there is NO $\delta>0$ with the property that: 
if |$x-c$|<$\delta$, then |$f(x)-f(c)$|<$\epsilon_0$.
Now, since $f$ is increasing on [a,b], it is one-to-one on [a,b]. This allows us to look at the expression
$f(c)-\epsilon_0<f(c)<f(c)+\epsilon_0$, and realize that there must be an $x_1,x_2\in[a,b]$ such that $x_1<x_2$ and,
$f(x_1)=f(c)-\epsilon_0$
$f(x_2)=f(c)+\epsilon_0$
So we have $f(x_1)<f(c)<f(x_2)$, and since $f$ is increasing, $x_1<c<x_2$.
So now choose $\delta$=min{$x_2-c,c-x_1$}
Now with this $\delta$ in mind, if |$x-c$|<$\delta$, it must be the case that $f(x_1)=f(c)-\epsilon_0<f(x)<f(c)+\epsilon_0=f(x_2)$.
But we suppose that there wasn't any $\delta$ that would work for this $\epsilon_0$.
