First suppose $\lim s_n = 0$. then from the definition we know that for all $\epsilon>0$ there exist an $N\in\mathbb{N}$ such that for all $n\in\mathbb{N}$ with $n>N$ we have, \begin{align*} |s_n-0| = |s_n|<\epsilon. \end{align*} It follows then that for $\epsilon$, $N$, and $n$ as before we have, \begin{align*} |s_n| = ||s_n|-0|<\epsilon. \end{align*} So $\lim |s_n|=0$. It follows from a theorem discussed in class that $\limsup|s_n|=\lim|s_n|=0$.
Now conversely suppose that $\limsup |s_n| = 0$. We know that, \begin{align*} \liminf|s_n|\le\limsup|s_n| = 0. \end{align*} However $|s_n|\ge 0$ for all $n\in\mathbb{N}$. So it follows that $\liminf|s_n|=0$. Since $\liminf|s_n|=\limsup|s_n|=0$ it follows that $\lim|s_n|=0$. Now since $\lim|s_n|=0$, we know that for every $\epsilon>0$ there is an $N\in\mathbb{N}$, such that for all $n>N$ we have, \begin{align*} ||s_n|-0|=|s_n|<\epsilon. \end{align*} Which implies that for $\epsilon$, $N$, and $n$ as above we have \begin{align*} |s_n-0|<\epsilon. \end{align*} Hence $\lim s_n=0$ as needed.
indent Therefore $\limsup |s_n| = 0$ if and only if $\lim s_n=0$. Q.E.D.
Is this proof correct, if not can I get a hint as to whats wrong with it.