Is this short proof valid for $\lim_{n \to \infty} c^n = 0$ where $|c|<1$?
Proof:
Limits are unique, and therefore it is sufficient to find the limit for one value of c. Notice if $c=0$ then $c^n=0$ for any $n\in\mathbb{Z_+}$. Then given an arbitrary $\varepsilon>0$, $|a_n-l|<\varepsilon ,\,\,\forall n\in\mathbb{Z_+}$. Since the limit of a sequence is unique then $\lim_{n \to \infty} c^n = 0$.
I know it is not valid obviously because then if the condition were $\forall c$ instead of $|c|<1$ then it wouldn't make the correct result.But where am I wrong?
The following statement is indeed true:
In your proof you have assumed that the limit exists, but you have yet to show that it does. We don't know anything about the sequence $c^n$ until we prove it.
But even if you know that the limit exists, your proof is still wrong. The reason is that $c^n$ does not give the same sequence for every value of $c$. For example $c=0$ gives the sequence $(0)_{n=1}^\infty$ and $c=1/2$ gives the sequence $((1/2)^n)_{n=1}^\infty$. These two sequences are different.