How to prove the following theorem?
Let $I \in R$ be an open interval, let $c \in I$, and let $f: I-{c} \rightarrow R$ be a function. Then $\lim \limits_{x \rightarrow c} {f(x)}$ exists if for each $\epsilon > 0$, there is some $\delta > 0$ such that $x,y \in I$ and $\vert x-c \vert < \delta$ and $\vert y-c \vert < \delta$ implies $\vert f(x)-f(y) \vert < \epsilon$.
I think this needs standard limit definition, sup and inf properties to prove. And I came up with a following scratch of proof:
(1) Suppose for each $\epsilon >0$, there is some $\delta >0$ such that $\vert x-c \vert < \delta$ and $\vert y-c \vert < \delta$ implies $\vert f(x)-f(y) \vert < \epsilon$. For each $r>0$, let $A_r$ = $I \bigcap (c-r,c+r)$. Then, for each $\epsilon>0$, there is some $\delta>0$, such that $x,y \in A_\delta$ implies $\vert f(x)-f(y) \vert < \epsilon$. Then I want to show there is some $a>0$ such that $f(A_a)$ is bounded.
(2) If $f(A_a)$ is bounded, then for each $s \in (0,a)$, $f(A_s) \subseteq f(A_a)$, and thus $f(A_s)$ is bounded. Define $a_s = \mathrm{glb}~f(A_s)$ and $b_s = \mathrm{lub}~f(A_s)$. Let $A = \{a_s \mid s \in (0,a)\}$ and $B=\{b_s \mid s \in (0,a)\}$. Then we know that A has a least upper bound and B has a greatest lower bound, and $\mathrm{lub}(A) \leq \mathrm{glb}(B)$. Now I want to show $\mathrm{lub}(A) = \mathrm{glb}(B)$
(3) If I could show $\mathrm{lub}(A) = \mathrm{glb}(B)$, let $M=\mathrm{lub}(A) = \mathrm{glb}(B)$, I want to show $\lim \limits_{x \rightarrow c} {f(x)} = M$.
Can someone give me some help on how to prove (1) (2) (3), $f(A_a)$ is bounded, $\mathrm{lub}(A) = \mathrm{glb}(B)$, and $\lim \limits_{x \rightarrow c} {f(x)} = M$? Thanks!