extend an alternative definition of limits to one-sided limits The following theorem is an alternative definition of limits. In this theorem, you don't need to know the value of $\lim \limits_{x \rightarrow c} {f(x)}$ in order to prove the limit exists.

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 $x,y \in I-{c}$ and $\vert x-c \vert < \delta$ and $\vert y-c \vert < \delta$ implies $\vert f(x)-f(y) \vert < \epsilon$.

I have extended this to a theorem for one-sided limits: $\lim \limits_{x \rightarrow c+} {f(x)}$ exists if or each $\epsilon > 0$, there is some $\delta > 0$ such that $c<x<c+ \delta$, and $c<y<c+ \delta$ implies $\vert f(x)-f(y) \vert < \epsilon$.
My question is: how to prove this extension to one-sided limits? I'm not familiar with Cauchy sequences, so I would prefer $\epsilon$-$\delta$ proof. 

Assume that $\lim_{x \to c^+} f(x) = g$ exists. Then for any $\varepsilon > 0$ there exists $\delta > 0$ such that if $0 < x - c < \delta$ then $|f(x) - g| < \varepsilon$ and if $0 < y - c < \delta$ then $|f(y) - g| < \varepsilon$. Hence
$$
|f(x) - f(y)| \leq |f(x) - g| + |f(y) - g| < 2\varepsilon
$$
for $0 < x - c < \delta$ and $0 < y - c < \delta$.
On the other hand, assume that your definition holds. Then there exists a strictly decreasing sequence $(c_n)_{n \geq 1}$ such that $c_n \to c$. Thus for a given $\varepsilon > 0$ there exists $\delta > 0$ such that there exists positive integer $N$ such that $|f(c_m)-f(c_n)| < \varepsilon$ for $m,n > N$.
It means that $(f(c_n))$ is Cauchy sequence and converges to, say, $g$.
(If you are not familiar with Cauchy sequences, it is easy to prove that the sequence $(a_n)$ such that for $\varepsilon > 0$ there exists $N > 0$ such that $|a_m - a_n| < \varepsilon$ for $m,n > N$ converges. To prove it, notice first that $(a_n)$ is bounded. Then, by Bolzano-Weierstrass theorem there is a subsequence $(a_{n_k})$ which converges to some $a$. Then, by Cauchy condition and triangle inequality $a_n \to a$.)
Assume now that there exists a sequence $(c'_n)_{n \geq 1}$ such that $f(c'_n) \to g' \neq g$. Let $C_n$ equals $c_n$ if $n$ is even, and equals $c'_n$ if $n$ is odd. Hence $(f(C_n))$ diverges but on the other hand we have $|f(C_m) - f(C_n)| < \varepsilon$ for sufficiently large $m,n$ and this means that $(f(C_n))$ is Cauchy, hence converges. Contradiction.
