Proving a theorem about the limit of a function The theorem is as follows:
$$\exists L\in\mathbb R\ \bigg(\lim_{x \to c}f(x)=L\bigg)\iff\forall\varepsilon>0\ \exists\delta>0\ \forall x_1,x_2\in B_{\delta}(c):|f(x_1)-f(x_2)|<\varepsilon$$
where $B_{\delta}$ is the set $\{x\in \mathscr D_f \ | \ 0<|x-c|<\delta\}$
It is quite easy to prove the implication $LHS\Rightarrow RHS$.
It is the other one I am struggling with( $LHS\Leftarrow RHS$ ).I'd appreciate a sketch of the proof or a hint on how to approach this problem. Thanks.

Your right hand side is called Cauchy's criterion for (continuous) limits. 
A hint for the proof is the following:
The right hand side implies that for every sequence $(x_i)$ converging to $c$ (with $x_i \neq c$ for all i), the sequence $f(x_i)$ is Cauchy and hence convergent. 
By "mixing" two different sequences $(x_i)$ and $(y_i)$ you can show that the limit is independent of the sequence. Call it L and show that this implies that $f(x)$ converges to $L$ as $x \rightarrow c$. 
