How to prove Cauchy Criterion for limits

ad 1: As you saw yourself, the assumption that the $\lim$ doesn't exist is difficult to handle, since there are an unfinite number of values $\eta$ to which $f(x)$ doesn't converge. In fact we need some heavier tools to prove the claim.

If $f$ is unbounded in a neighborhood of $c$ then construct a sequence $(x_n)_{n\geq0}$ recursively as follows: Choose $x_0$ arbitrarily, and for each $n\geq1$ choose an $x_n$ with $|x_n-c|<{1\over n}$ and $|f(x_n)|\geq |f(x_{n-1})|+1$. Then the $x_n$ converge to $c$, as do the $y_n:=x_{n-1}$ $\>(n\geq1)$. Furthermore we have $|f(x_n)-f(y_n)|\geq1$ for all $n\geq1$.

When $f$ is bounded choose an arbitrary sequence $(x_n)_{n\geq0}$ converging to $c$. The sequence $a_n:=f(x_n)$ is a bounded sequence of real numbers and so has an accumulation point $\alpha\in{\mathbb R}$. Passing to a subsequence we may assume $\lim_{n\to\infty} f(x_n)=\alpha$. As, by assumption, the $\lim_{x\to c} f(x)$ does not exist there is an $\epsilon>0$ and a sequence $(y_n)_{n\geq0}$ converging to $c$ with $|f(y_n)-\alpha|\geq\epsilon$ for all $n$. It follows that we have $|f(x_n)-f(y_n)|\geq{\epsilon\over2}$ for all large enough $n$.

A second way to prove 1. would be to prove 2. first.

ad 2: This is the "Cauchy criterion for functions". As with the Cauchy criterion for sequences one direction is trivial. So assume that for all $\epsilon>0$ there is a $\delta>0$ such that $\ldots$

Choose an arbitrary sequence $(x_n)_{n\geq0}$ converging to $c$. Then it is easy to see that $\bigl(f(x_n)\bigr)_{n\geq0}$ is a Cauchy sequence in ${\mathbb R}$ and therefore converges to a number $\alpha\in{\mathbb R}$. Given an $\epsilon>0$ there is a $\delta>0$ with $$|f(y)-f(x_n)|<\epsilon\tag{1}$$ for all $y\in \dot U_\delta(c)$ and all large enough $n$. For fixed $y$ letting $n\to\infty$ in $(1)$ we conclude that $$|f(y)-\alpha|\leq\epsilon\quad\forall \ y\in \dot U_\delta(c)\ .$$