You have essentially the same difficulty as in the previous question: figuring out a "target" number $L$ for the limit.
(By the way: if you don't know what Cauchy sequences are, you should have said something in that previous problem!)
You can proceed in a similar manner. First, you want to find a potential "target." For $\epsilon_n = \frac{1}{n}$, you know there is a $\delta_n$ (which you may assume is less than or equal to $\frac{1}{n}$ and less than $\delta_{n-1}$) such that for all $x$ and $y$, if $c\lt x\lt c+\delta_n$ and $c\lt y \lt c+\delta_n$, then $|f(x)-f(y)|\lt \frac{1}{n}$.
Note that this means that the values of $f$ on $(c,c+\delta_n)$ are bounded: for any $x\in (c,c+\delta_n)$, you know that $|f(x) - f(c+\delta_n/2)|\lt \frac{1}{n}$, so $|f(x)| \lt \frac{1}{n}+|f(c+\delta_n/2)|$.
In particular, there is a greatest lower bound $a_n$ to $f(c,c+\delta_n)$. Since $(c,c+\delta_{n+1})\subseteq (c,c+\delta_n)$, we have that $a_n\leq a_{n+1}$. So the sequence $a_1,a_2,\ldots$ is an increasing function. The sequence is bounded above, so the sequence converges to some $L$.
The obvious "target" for the limit is $L$. So, let $\epsilon\gt 0$, and you want to show that there is a $\delta\gt 0$ such that for all $x$, if $c\lt x\lt c+\delta$, then $|f(x)-L|\lt \epsilon$.
Since $L$ is the least upper bound of $a_1,a_2,\ldots$, there exists $N$ such that $L-\frac{\epsilon}{2} \lt a_n \leq L$ for all $n\geq N$. And there exists an $M$ such that $\frac{1}{M}\lt \frac{\epsilon}{4}$. Let $K=\max\{M,N\}$, and consider $\delta=\delta_K$.
We know that since $a_K$ is the greatest lower bound of $f(c,c+\delta_K)$, there exists $y\in (c,c+\delta_K)$ such that $a_K\leq f(y)\lt a_K+\frac{\epsilon}{4}$.
Now suppose that $c\lt x \lt c+\delta_K$. Then \begin{align*} |f(x)-L| &\leq |f(x)-f(y)|+|f(y)-a_K| + |a_K-L| &&\mbox{(triangle inequality)}\\\ &\lt \frac{1}{K} + |f(y)-a_K| + |a_K-L| &&\mbox{(choice of $\delta_K$)}\\\ &\leq \frac{1}{K} + \frac{\epsilon}{4} + |a_K-L| &&\mbox{(choice of $y$)}\\\ &\leq \frac{1}{K} + \frac{\epsilon}{4} + \frac{\epsilon}{2} &&\mbox{(since $K\geq N$)}\\\ &\lt \frac{\epsilon}{4} + \frac{\epsilon}{4} + \frac{\epsilon}{2} &&\mbox{(since $K\geq M$)}\\\ &=\epsilon. \end{align*}
(With this in mind, you can try doing the two-sided version of the problem without invoking Cauchy sequences; the idea is very similar, and you don't need to consider both least upper bounds and greatest lower bounds for $f(c-\delta_n,c+\delta_n)$, just one of the two and use their limit as a "target". Once you know what to aim for, hitting it becomes much easier).