If $(s_n)$ converges to $s$, then $(\lvert s_n\rvert)$ converges to $\lvert s\rvert$. Question:

If $(s_n)$ converges to $s$, then $(\lvert s_n\rvert)$ converges to $\lvert s\rvert$. Prove or give a counterexample.


Attempt:

The statement is true because if $(\lvert s_n\rvert)$ converges to $\lvert s\rvert$, then given any $\epsilon >0$ there exists an $N\in\mathbb{N}$ such that for all $n\in\mathbb{N}$ with $n\geq N$ we have that $\lvert \lvert s_n\rvert-\lvert s\rvert\rvert <\epsilon$, but $\lvert \lvert s_n\rvert-\lvert s\rvert\rvert \leq\lvert s_n-s\rvert$ by the reverse triangle inequality, and by assumption given any $\epsilon >0$ there exists an $N\in\mathbb{N}$ such that for all $n\in\mathbb{N}$ with $n\geq N$ we have that $\lvert s_n-s\rvert <\epsilon$; hence
  $$\lvert \lvert s_n\rvert -\lvert s\rvert \rvert\leq \lvert s_n-s\rvert<\epsilon.$$


From the hypothesis you know that fixed any $\varepsilon >0$ there exists $N \in \mathbb N$ s.t. $n>N$ implies $\vert s-s_n \vert <\varepsilon$. 
By reverse triangular inequality, you get that
$$
\vert \vert s_n \vert - \vert s \vert \vert \le \vert s-s_n\vert < \varepsilon
$$
if $n >N$. This is equivalent to say $\vert s_n \vert \to \vert s \vert$, which is the claim. Hope this can help. 
