Recall that $\lim_{x\to a}\ f(x)=L$ means that for all $\epsilon>0$ we are able to find a $\delta>0$ such that if $0<|x-a|<\delta$ then $|f(x)-L|<\epsilon$.

In the case of $\lim_{x\to a}\ c=c$ this means that for all $\epsilon>0$ we have to find a $\delta>0$ such that if $0<|x-a|<\delta$ then $|c-c|<\epsilon$.

So let $\epsilon$ be any positive real number. Then the condition $|c-c|<\epsilon$ is always true since we are assuming that $\epsilon>0$.

Therefore we can choose $\delta$ to be any positive real number, e.g. $\delta=1$.

With this (or any other) choice of $\delta$, it is true that if $|x-a|<1$ then $|c-c|<\epsilon$ (since the conclusion is always true).

This proves that $\lim_{x\to a}\ c=c$.