Is the following argument correct?
For a given sequence $(s_n)$ show that $\lim s_n = 0$ if and only if $\lim |s_n| = 0$.
Proof. $(\Rightarrow).$ Let $\epsilon>0$ since $\lim s_n = 0$ we may invoke an $N\in\mathbf{R}^+$ such that given any $k\in\mathbf{Z^+}$ we have $|s_k|<\epsilon$ whenever $k>N$ but $|s_k| = ||s_k|-0|$ consequently $||s_k|-0|<\epsilon$ implying that $\lim |s_n| = 0$.
$(\Leftarrow).$ Conversely assume now that $\lim |s_n| = 0$ but $\lim s_n\neq 0$. Since $\lim s_n\neq 0$ it follows that for some $\epsilon>0$ given any $M\in\mathbf{R}^+$ there exist $k\in\mathbf{Z^+}$ such that $k>M$ and $|a_k|\ge\epsilon$ but $\lim |s_n| = 0$ consequently for some $N\in\mathbf{R}^+$ given any $n\in\mathbf{Z}^+$ we have $||a_n|-0|<\epsilon$ whenever $n>N$ then in particular for $M=N$ we have $|a_k|\ge\epsilon$ and since $|a_k| = ||a_k|-0|$ it follows that $||a_k|-0|\ge\epsilon$ but $k>N$ and so $||a_k|-0|<\epsilon$ resulting in a contradiction.
Ok basically but much too complicated ,Particularly the reverse direction .Look, $|s_n-0|<\epsilon$ is the same as $||s_n|-0|<\epsilon$ namely $|s_n|<\epsilon$ so the two definitions of limits coincide .